Solve 6=x^2+3x-48

$$6 = x ^ { 2 } + 3 x - 48$$

Solve for x

$$x=-9$$

$$x=6$$

Steps Using Factoring

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

$$x^{2}+3x-48=6$$

Subtract 6 from both sides.

$$x^{2}+3x-48-6=0$$

Subtract 6 from -48 to get -54.

$$x^{2}+3x-54=0$$

To solve the equation, factor x^{2}+3x-54 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.

$$a+b=3 ab=-54$$

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 -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 3.

$$a=-6 b=9$$

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

$$\left(x-6\right)\left(x+9\right)$$

To find equation solutions, solve x-6=0 and x+9=0.

$$x=6 x=-9$$

Steps Using Factoring By Grouping

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

$$x^{2}+3x-48=6$$

Subtract 6 from both sides.

$$x^{2}+3x-48-6=0$$

Subtract 6 from -48 to get -54.

$$x^{2}+3x-54=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-54. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-54\right)=-54$$

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 -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 3.

$$a=-6 b=9$$

Rewrite x^{2}+3x-54 as \left(x^{2}-6x\right)+\left(9x-54\right).

$$\left(x^{2}-6x\right)+\left(9x-54\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x-6\right)+9\left(x-6\right)$$

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

$$\left(x-6\right)\left(x+9\right)$$

To find equation solutions, solve x-6=0 and x+9=0.

$$x=6 x=-9$$

Steps Using the Quadratic Formula

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

$$x^{2}+3x-48=6$$

Subtract 6 from both sides.

$$x^{2}+3x-48-6=0$$

Subtract 6 from -48 to get -54.

$$x^{2}+3x-54=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-54\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-54\right)}}{2}$$

Multiply -4 times -54.

$$x=\frac{-3±\sqrt{9+216}}{2}$$

Add 9 to 216.

$$x=\frac{-3±\sqrt{225}}{2}$$

Take the square root of 225.

$$x=\frac{-3±15}{2}$$

Now solve the equation x=\frac{-3±15}{2} when ± is plus. Add -3 to 15.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{-3±15}{2} when ± is minus. Subtract 15 from -3.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=6 x=-9$$

Steps for Completing the Square

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

$$x^{2}+3x-48=6$$

Add 48 to both sides.

$$x^{2}+3x=6+48$$

Add 6 and 48 to get 54.

$$x^{2}+3x=54$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=54+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=54+\frac{9}{4}$$

Add 54 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{225}{4}$$

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

$$\left(x+\frac{3}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{15}{2} x+\frac{3}{2}=-\frac{15}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=6 x=-9$$