Solve 0=2x^2+3x-90

$$0 = 2 { x }^{ 2 } +3x-90$$

Solve for x

$$x = -\frac{15}{2} = -7\frac{1}{2} = -7.5$$

$$x=6$$

Steps Using Factoring By Grouping

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

$$2x^{2}+3x-90=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-90. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-90\right)=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 3.

$$a=-12 b=15$$

Rewrite 2x^{2}+3x-90 as \left(2x^{2}-12x\right)+\left(15x-90\right).

$$\left(2x^{2}-12x\right)+\left(15x-90\right)$$

Factor out 2x in the first and 15 in the second group.

$$2x\left(x-6\right)+15\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(2x+15\right)$$

To find equation solutions, solve x-6=0 and 2x+15=0.

$$x=6 x=-\frac{15}{2}$$

Steps Using the Quadratic Formula

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

$$2x^{2}+3x-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-90\right)}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-90\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-90\right)}}{2\times 2}$$

Multiply -8 times -90.

$$x=\frac{-3±\sqrt{9+720}}{2\times 2}$$

Add 9 to 720.

$$x=\frac{-3±\sqrt{729}}{2\times 2}$$

Take the square root of 729.

$$x=\frac{-3±27}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±27}{4}$$

Now solve the equation x=\frac{-3±27}{4} when ± is plus. Add -3 to 27.

$$x=\frac{24}{4}$$

Divide 24 by 4.

$$x=6$$

Now solve the equation x=\frac{-3±27}{4} when ± is minus. Subtract 27 from -3.

$$x=-\frac{30}{4}$$

Reduce the fraction \frac{-30}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=6 x=-\frac{15}{2}$$

Steps for Completing the Square

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

$$2x^{2}+3x-90=0$$

Add 90 to both sides. Anything plus zero gives itself.

$$2x^{2}+3x=90$$

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{90}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{90}{2}$$

Divide 90 by 2.

$$x^{2}+\frac{3}{2}x=45$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=45+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=45+\frac{9}{16}$$

Add 45 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{729}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{729}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{729}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{27}{4} x+\frac{3}{4}=-\frac{27}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=6 x=-\frac{15}{2}$$