Solve -2x^2=7x-15

$$- 2 x ^ { 2 } = 7 x - 15$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=-5$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

$$-2x^{2}-7x=-15$$

Add 15 to both sides.

$$-2x^{2}-7x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-2\times 15=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=3 b=-10$$

Rewrite -2x^{2}-7x+15 as \left(-2x^{2}+3x\right)+\left(-10x+15\right).

$$\left(-2x^{2}+3x\right)+\left(-10x+15\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(2x-3\right)-5\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-x-5\right)$$

To find equation solutions, solve 2x-3=0 and -x-5=0.

$$x=\frac{3}{2} x=-5$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$-2x^{2}-7x=-15$$

Add 15 to both sides.

$$-2x^{2}-7x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -7 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-2\right)\times 15}}{2\left(-2\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-2\right)\times 15}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-7\right)±\sqrt{49+8\times 15}}{2\left(-2\right)}$$

Multiply 8 times 15.

$$x=\frac{-\left(-7\right)±\sqrt{49+120}}{2\left(-2\right)}$$

Add 49 to 120.

$$x=\frac{-\left(-7\right)±\sqrt{169}}{2\left(-2\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-7\right)±13}{2\left(-2\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±13}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{7±13}{-4}$$

Now solve the equation x=\frac{7±13}{-4} when ± is plus. Add 7 to 13.

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

Divide 20 by -4.

$$x=-5$$

Now solve the equation x=\frac{7±13}{-4} when ± is minus. Subtract 13 from 7.

$$x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=-5 x=\frac{3}{2}$$

Steps for Completing the Square

Subtract 7x from both sides.

$$-2x^{2}-7x=-15$$

Divide both sides by -2.

$$\frac{-2x^{2}-7x}{-2}=-\frac{15}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{7}{-2}\right)x=-\frac{15}{-2}$$

Divide -7 by -2.

$$x^{2}+\frac{7}{2}x=-\frac{15}{-2}$$

Divide -15 by -2.

$$x^{2}+\frac{7}{2}x=\frac{15}{2}$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=\frac{15}{2}+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$$

Add \frac{15}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{169}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{13}{4} x+\frac{7}{4}=-\frac{13}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{3}{2} x=-5$$