Solve 2x^2=15x+8

$$2 x ^ { 2 } = 15 x + 8$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=8$$

Steps Using Factoring By Grouping

Subtract 15x from both sides.

$$2x^{2}-15x=8$$

Subtract 8 from both sides.

$$2x^{2}-15x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-15 ab=2\left(-8\right)=-16$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -16.

$$1,-16 2,-8 4,-4$$

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -15.

$$a=-16 b=1$$

Rewrite 2x^{2}-15x-8 as \left(2x^{2}-16x\right)+\left(x-8\right).

$$\left(2x^{2}-16x\right)+\left(x-8\right)$$

Factor out 2x in 2x^{2}-16x.

$$2x\left(x-8\right)+x-8$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(2x+1\right)$$

To find equation solutions, solve x-8=0 and 2x+1=0.

$$x=8 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 15x from both sides.

$$2x^{2}-15x=8$$

Subtract 8 from both sides.

$$2x^{2}-15x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -15 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-\left(-15\right)±\sqrt{225+64}}{2\times 2}$$

Add 225 to 64.

$$x=\frac{-\left(-15\right)±\sqrt{289}}{2\times 2}$$

Take the square root of 289.

$$x=\frac{-\left(-15\right)±17}{2\times 2}$$

The opposite of -15 is 15.

$$x=\frac{15±17}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{15±17}{4}$$

Now solve the equation x=\frac{15±17}{4} when ± is plus. Add 15 to 17.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{15±17}{4} when ± is minus. Subtract 17 from 15.

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

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

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=8 x=-\frac{1}{2}$$

Steps for Completing the Square

Subtract 15x from both sides.

$$2x^{2}-15x=8$$

Divide both sides by 2.

$$\frac{2x^{2}-15x}{2}=\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{15}{2}x=\frac{8}{2}$$

Divide 8 by 2.

$$x^{2}-\frac{15}{2}x=4$$

Divide -\frac{15}{2}, the coefficient of the x term, by 2 to get -\frac{15}{4}. Then add the square of -\frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=4+\left(-\frac{15}{4}\right)^{2}$$

Square -\frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=4+\frac{225}{16}$$

Add 4 to \frac{225}{16}.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{289}{16}$$

Factor x^{2}-\frac{15}{2}x+\frac{225}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{289}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{289}{16}}$$

Simplify.

$$x-\frac{15}{4}=\frac{17}{4} x-\frac{15}{4}=-\frac{17}{4}$$

Add \frac{15}{4} to both sides of the equation.

$$x=8 x=-\frac{1}{2}$$