Solve 0=x^2+7x-30

$$0 = { x }^{ 2 } +7x-30$$

Solve for x

$$x=-10$$

$$x=3$$

Steps Using Factoring

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

$$x^{2}+7x-30=0$$

To solve the equation, factor x^{2}+7x-30 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=7 ab=-30$$

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 -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 factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+10\right)$$

To find equation solutions, solve x-3=0 and x+10=0.

$$x=3 x=-10$$

Steps Using Factoring By Grouping

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

$$x^{2}+7x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-30\right)=-30$$

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 -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 x^{2}+7x-30 as \left(x^{2}-3x\right)+\left(10x-30\right).

$$\left(x^{2}-3x\right)+\left(10x-30\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-3\right)+10\left(x-3\right)$$

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

$$\left(x-3\right)\left(x+10\right)$$

To find equation solutions, solve x-3=0 and x+10=0.

$$x=3 x=-10$$

Steps Using the Quadratic Formula

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

$$x^{2}+7x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-30\right)}}{2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-7±\sqrt{49+120}}{2}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-7±13}{2}$$

Now solve the equation x=\frac{-7±13}{2} when ± is plus. Add -7 to 13.

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

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-7±13}{2} when ± is minus. Subtract 13 from -7.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=3 x=-10$$

Steps for Completing the Square

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

$$x^{2}+7x-30=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$x^{2}+7x=30$$

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

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=30+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=30+\frac{49}{4}$$

Add 30 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{169}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{7}{2}=\frac{13}{2} x+\frac{7}{2}=-\frac{13}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=3 x=-10$$