Solve (3x - 1)(72 * 45) * 70 = (6x + 2) ^ 2 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$(3x-1)(72\times45)70=(6x+2)^{2}$$

Solve for x

$x=30\sqrt{99211}+\frac{28349}{3}\approx 18898.999976483$
$x=\frac{28349}{3}-30\sqrt{99211}\approx 0.333356851$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply $72$ and $45$ to get $3240$.

$$\left(3x-1\right)\times 3240\times 70=\left(6x+2\right)^{2}$$

Multiply $3240$ and $70$ to get $226800$.

$$\left(3x-1\right)\times 226800=\left(6x+2\right)^{2}$$

Use the distributive property to multiply $3x-1$ by $226800$.

$$680400x-226800=\left(6x+2\right)^{2}$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(6x+2\right)^{2}$.

$$680400x-226800=36x^{2}+24x+4$$

Subtract $36x^{2}$ from both sides.

$$680400x-226800-36x^{2}=24x+4$$

Subtract $24x$ from both sides.

$$680400x-226800-36x^{2}-24x=4$$

Combine $680400x$ and $-24x$ to get $680376x$.

$$680376x-226800-36x^{2}=4$$

Subtract $4$ from both sides.

$$680376x-226800-36x^{2}-4=0$$

Subtract $4$ from $-226800$ to get $-226804$.

$$680376x-226804-36x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-36x^{2}+680376x-226804=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-36$ for $a$, $680376$ for $b$, and $-226804$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-680376±\sqrt{680376^{2}-4\left(-36\right)\left(-226804\right)}}{2\left(-36\right)}$$

Square $680376$.

$$x=\frac{-680376±\sqrt{462911501376-4\left(-36\right)\left(-226804\right)}}{2\left(-36\right)}$$

Multiply $-4$ times $-36$.

$$x=\frac{-680376±\sqrt{462911501376+144\left(-226804\right)}}{2\left(-36\right)}$$

Multiply $144$ times $-226804$.

$$x=\frac{-680376±\sqrt{462911501376-32659776}}{2\left(-36\right)}$$

Add $462911501376$ to $-32659776$.

$$x=\frac{-680376±\sqrt{462878841600}}{2\left(-36\right)}$$

Take the square root of $462878841600$.

$$x=\frac{-680376±2160\sqrt{99211}}{2\left(-36\right)}$$

Multiply $2$ times $-36$.

$$x=\frac{-680376±2160\sqrt{99211}}{-72}$$

Now solve the equation $x=\frac{-680376±2160\sqrt{99211}}{-72}$ when $±$ is plus. Add $-680376$ to $2160\sqrt{99211}$.

$$x=\frac{2160\sqrt{99211}-680376}{-72}$$

Divide $-680376+2160\sqrt{99211}$ by $-72$.

$$x=\frac{28349}{3}-30\sqrt{99211}$$

Now solve the equation $x=\frac{-680376±2160\sqrt{99211}}{-72}$ when $±$ is minus. Subtract $2160\sqrt{99211}$ from $-680376$.

$$x=\frac{-2160\sqrt{99211}-680376}{-72}$$

Divide $-680376-2160\sqrt{99211}$ by $-72$.

$$x=30\sqrt{99211}+\frac{28349}{3}$$

The equation is now solved.

$$x=\frac{28349}{3}-30\sqrt{99211}$$ $$x=30\sqrt{99211}+\frac{28349}{3}$$

Steps for Completing the Square

Multiply $72$ and $45$ to get $3240$.

$$\left(3x-1\right)\times 3240\times 70=\left(6x+2\right)^{2}$$

Multiply $3240$ and $70$ to get $226800$.

$$\left(3x-1\right)\times 226800=\left(6x+2\right)^{2}$$

Use the distributive property to multiply $3x-1$ by $226800$.

$$680400x-226800=\left(6x+2\right)^{2}$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(6x+2\right)^{2}$.

$$680400x-226800=36x^{2}+24x+4$$

Subtract $36x^{2}$ from both sides.

$$680400x-226800-36x^{2}=24x+4$$

Subtract $24x$ from both sides.

$$680400x-226800-36x^{2}-24x=4$$

Combine $680400x$ and $-24x$ to get $680376x$.

$$680376x-226800-36x^{2}=4$$

Add $226800$ to both sides.

$$680376x-36x^{2}=4+226800$$

Add $4$ and $226800$ to get $226804$.

$$680376x-36x^{2}=226804$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-36x^{2}+680376x=226804$$

Divide both sides by $-36$.

$$\frac{-36x^{2}+680376x}{-36}=\frac{226804}{-36}$$

Dividing by $-36$ undoes the multiplication by $-36$.

$$x^{2}+\frac{680376}{-36}x=\frac{226804}{-36}$$

Reduce the fraction $\frac{680376}{-36}$ to lowest terms by extracting and canceling out $12$.

$$x^{2}-\frac{56698}{3}x=\frac{226804}{-36}$$

Reduce the fraction $\frac{226804}{-36}$ to lowest terms by extracting and canceling out $4$.

$$x^{2}-\frac{56698}{3}x=-\frac{56701}{9}$$

Divide $-\frac{56698}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{28349}{3}$. Then add the square of $-\frac{28349}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{56698}{3}x+\left(-\frac{28349}{3}\right)^{2}=-\frac{56701}{9}+\left(-\frac{28349}{3}\right)^{2}$$

Square $-\frac{28349}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{56698}{3}x+\frac{803665801}{9}=\frac{-56701+803665801}{9}$$

Add $-\frac{56701}{9}$ to $\frac{803665801}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{56698}{3}x+\frac{803665801}{9}=89289900$$

Factor $x^{2}-\frac{56698}{3}x+\frac{803665801}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{28349}{3}\right)^{2}=89289900$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{28349}{3}\right)^{2}}=\sqrt{89289900}$$

Simplify.

$$x-\frac{28349}{3}=30\sqrt{99211}$$ $$x-\frac{28349}{3}=-30\sqrt{99211}$$

Add $\frac{28349}{3}$ to both sides of the equation.

$$x=30\sqrt{99211}+\frac{28349}{3}$$ $$x=\frac{28349}{3}-30\sqrt{99211}$$