Solve (x^2+3x-7)+(6x-5)

$$( x ^ { 2 } + 3 x - 7 ) + ( 6 x - 5 )$$

Evaluate

$$x^{2}+9x-12$$

Solution Steps

Combine 3x and 6x to get 9x.

$$x^{2}+9x-7-5$$

Subtract 5 from -7 to get -12.

$$x^{2}+9x-12$$

Factor

$$\left(x-\frac{-\sqrt{129}-9}{2}\right)\left(x-\frac{\sqrt{129}-9}{2}\right)$$

Steps Using the Quadratic Formula

Combine 3x and 6x to get 9x.

$$factor(x^{2}+9x-7-5)$$

Subtract 5 from -7 to get -12.

$$factor(x^{2}+9x-12)$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+9x-12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-12\right)}}{2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-9±\sqrt{81+48}}{2}$$

Add 81 to 48.

$$x=\frac{-9±\sqrt{129}}{2}$$

Now solve the equation x=\frac{-9±\sqrt{129}}{2} when ± is plus. Add -9 to \sqrt{129}.

$$x=\frac{\sqrt{129}-9}{2}$$

Now solve the equation x=\frac{-9±\sqrt{129}}{2} when ± is minus. Subtract \sqrt{129} from -9.

$$x=\frac{-\sqrt{129}-9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-9+\sqrt{129}}{2} for x_{1} and \frac{-9-\sqrt{129}}{2} for x_{2}.

$$x^{2}+9x-12=\left(x-\frac{\sqrt{129}-9}{2}\right)\left(x-\frac{-\sqrt{129}-9}{2}\right)$$