Solve - 105 = k/2 * (32 - 2k) | 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

$$-105=\frac{k}{2}(32-2k)$$

Solve for k

$k=-5$
$k=21$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $2$.

$$-210=k\left(32-2k\right)$$

Use the distributive property to multiply $k$ by $32-2k$.

$$-210=32k-2k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$32k-2k^{2}=-210$$

Add $210$ to both sides.

$$32k-2k^{2}+210=0$$

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.

$$-2k^{2}+32k+210=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $32$ for $b$, and $210$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$k=\frac{-32±\sqrt{32^{2}-4\left(-2\right)\times 210}}{2\left(-2\right)}$$

Square $32$.

$$k=\frac{-32±\sqrt{1024-4\left(-2\right)\times 210}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$k=\frac{-32±\sqrt{1024+8\times 210}}{2\left(-2\right)}$$

Multiply $8$ times $210$.

$$k=\frac{-32±\sqrt{1024+1680}}{2\left(-2\right)}$$

Add $1024$ to $1680$.

$$k=\frac{-32±\sqrt{2704}}{2\left(-2\right)}$$

Take the square root of $2704$.

$$k=\frac{-32±52}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$k=\frac{-32±52}{-4}$$

Now solve the equation $k=\frac{-32±52}{-4}$ when $±$ is plus. Add $-32$ to $52$.

$$k=\frac{20}{-4}$$

Divide $20$ by $-4$.

$$k=-5$$

Now solve the equation $k=\frac{-32±52}{-4}$ when $±$ is minus. Subtract $52$ from $-32$.

$$k=-\frac{84}{-4}$$

Divide $-84$ by $-4$.

$$k=21$$

The equation is now solved.

$$k=-5$$ $$k=21$$

Steps for Completing the Square

Multiply both sides of the equation by $2$.

$$-210=k\left(32-2k\right)$$

Use the distributive property to multiply $k$ by $32-2k$.

$$-210=32k-2k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$32k-2k^{2}=-210$$

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 $x^{2}+bx=c$.

$$-2k^{2}+32k=-210$$

Divide both sides by $-2$.

$$\frac{-2k^{2}+32k}{-2}=-\frac{210}{-2}$$

Dividing by $-2$ undoes the multiplication by $-2$.

$$k^{2}+\frac{32}{-2}k=-\frac{210}{-2}$$

Divide $32$ by $-2$.

$$k^{2}-16k=-\frac{210}{-2}$$

Divide $-210$ by $-2$.

$$k^{2}-16k=105$$

Divide $-16$, the coefficient of the $x$ term, by $2$ to get $-8$. Then add the square of $-8$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-16k+\left(-8\right)^{2}=105+\left(-8\right)^{2}$$

Square $-8$.

$$k^{2}-16k+64=105+64$$

Add $105$ to $64$.

$$k^{2}-16k+64=169$$

Factor $k^{2}-16k+64$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(k-8\right)^{2}=169$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-8\right)^{2}}=\sqrt{169}$$

Simplify.

$$k-8=13$$ $$k-8=-13$$

Add $8$ to both sides of the equation.

$$k=21$$ $$k=-5$$