Solve 0=-5x^2+5x+30

$$0 = - 5 x ^ { 2 } + 5 x + 30$$

Solve for x

$$x=-2$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$-5x^{2}+5x+30=0$$

Divide both sides by 5.

$$-x^{2}+x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=1 ab=-6=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -x^{2}+x+6 as \left(-x^{2}+3x\right)+\left(-2x+6\right).

$$\left(-x^{2}+3x\right)+\left(-2x+6\right)$$

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

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

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

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

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

$$x=3 x=-2$$

Steps Using the Quadratic Formula

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

$$-5x^{2}+5x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 5 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-5\right)\times 30}}{2\left(-5\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-5\right)\times 30}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-5±\sqrt{25+20\times 30}}{2\left(-5\right)}$$

Multiply 20 times 30.

$$x=\frac{-5±\sqrt{25+600}}{2\left(-5\right)}$$

Add 25 to 600.

$$x=\frac{-5±\sqrt{625}}{2\left(-5\right)}$$

Take the square root of 625.

$$x=\frac{-5±25}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-5±25}{-10}$$

Now solve the equation x=\frac{-5±25}{-10} when ± is plus. Add -5 to 25.

$$x=\frac{20}{-10}$$

Divide 20 by -10.

$$x=-2$$

Now solve the equation x=\frac{-5±25}{-10} when ± is minus. Subtract 25 from -5.

$$x=-\frac{30}{-10}$$

Divide -30 by -10.

$$x=3$$

The equation is now solved.

$$x=-2 x=3$$

Steps for Completing the Square

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

$$-5x^{2}+5x+30=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}+5x=-30$$

Divide both sides by -5.

$$\frac{-5x^{2}+5x}{-5}=-\frac{30}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{5}{-5}x=-\frac{30}{-5}$$

Divide 5 by -5.

$$x^{2}-x=-\frac{30}{-5}$$

Divide -30 by -5.

$$x^{2}-x=6$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$