Solve 4x ^ 2 + 20 = 18 + 35x | 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

$$4x^{2}+20=18+35x$$

Solve for x

$x = \frac{\sqrt{1193} + 35}{8} \approx 8.69247901$
$x=\frac{35-\sqrt{1193}}{8}\approx 0.05752099$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $18$ from both sides.

$$4x^{2}+20-18=35x$$

Subtract $18$ from $20$ to get $2$.

$$4x^{2}+2=35x$$

Subtract $35x$ from both sides.

$$4x^{2}+2-35x=0$$

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.

$$4x^{2}-35x+2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $-35$ for $b$, and $2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 4\times 2}}{2\times 4}$$

Square $-35$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 4\times 2}}{2\times 4}$$

Multiply $-4$ times $4$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-16\times 2}}{2\times 4}$$

Multiply $-16$ times $2$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-32}}{2\times 4}$$

Add $1225$ to $-32$.

$$x=\frac{-\left(-35\right)±\sqrt{1193}}{2\times 4}$$

The opposite of $-35$ is $35$.

$$x=\frac{35±\sqrt{1193}}{2\times 4}$$

Multiply $2$ times $4$.

$$x=\frac{35±\sqrt{1193}}{8}$$

Now solve the equation $x=\frac{35±\sqrt{1193}}{8}$ when $±$ is plus. Add $35$ to $\sqrt{1193}$.

$$x=\frac{\sqrt{1193}+35}{8}$$

Now solve the equation $x=\frac{35±\sqrt{1193}}{8}$ when $±$ is minus. Subtract $\sqrt{1193}$ from $35$.

$$x=\frac{35-\sqrt{1193}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{1193}+35}{8}$$ $$x=\frac{35-\sqrt{1193}}{8}$$

Steps for Completing the Square

Subtract $35x$ from both sides.

$$4x^{2}+20-35x=18$$

Subtract $20$ from both sides.

$$4x^{2}-35x=18-20$$

Subtract $20$ from $18$ to get $-2$.

$$4x^{2}-35x=-2$$

Divide both sides by $4$.

$$\frac{4x^{2}-35x}{4}=-\frac{2}{4}$$

Dividing by $4$ undoes the multiplication by $4$.

$$x^{2}-\frac{35}{4}x=-\frac{2}{4}$$

Reduce the fraction $\frac{-2}{4}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{35}{4}x=-\frac{1}{2}$$

Divide $-\frac{35}{4}$, the coefficient of the $x$ term, by $2$ to get $-\frac{35}{8}$. Then add the square of $-\frac{35}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{4}x+\left(-\frac{35}{8}\right)^{2}=-\frac{1}{2}+\left(-\frac{35}{8}\right)^{2}$$

Square $-\frac{35}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{4}x+\frac{1225}{64}=-\frac{1}{2}+\frac{1225}{64}$$

Add $-\frac{1}{2}$ to $\frac{1225}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{35}{4}x+\frac{1225}{64}=\frac{1193}{64}$$

Factor $x^{2}-\frac{35}{4}x+\frac{1225}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{35}{8}\right)^{2}=\frac{1193}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{8}\right)^{2}}=\sqrt{\frac{1193}{64}}$$

Simplify.

$$x-\frac{35}{8}=\frac{\sqrt{1193}}{8}$$ $$x-\frac{35}{8}=-\frac{\sqrt{1193}}{8}$$

Add $\frac{35}{8}$ to both sides of the equation.

$$x=\frac{\sqrt{1193}+35}{8}$$ $$x=\frac{35-\sqrt{1193}}{8}$$