Solve (4x^2-6x+29)+(4x^2+9x-68)

$$( 4 x ^ { 2 } - 6 x + 29 ) + ( 4 x ^ { 2 } + 9 x - 68 )$$

Evaluate

$$8x^{2}+3x-39$$

Solution Steps

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}-6x+29+9x-68$$

Combine -6x and 9x to get 3x.

$$8x^{2}+3x+29-68$$

Subtract 68 from 29 to get -39.

$$8x^{2}+3x-39$$

Factor

$$8\left(x-\frac{-\sqrt{1257}-3}{16}\right)\left(x-\frac{\sqrt{1257}-3}{16}\right)$$

Steps Using the Quadratic Formula

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$factor(8x^{2}-6x+29+9x-68)$$

Combine -6x and 9x to get 3x.

$$factor(8x^{2}+3x+29-68)$$

Subtract 68 from 29 to get -39.

$$factor(8x^{2}+3x-39)$$

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.

$$8x^{2}+3x-39=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 8\left(-39\right)}}{2\times 8}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 8\left(-39\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-3±\sqrt{9-32\left(-39\right)}}{2\times 8}$$

Multiply -32 times -39.

$$x=\frac{-3±\sqrt{9+1248}}{2\times 8}$$

Add 9 to 1248.

$$x=\frac{-3±\sqrt{1257}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-3±\sqrt{1257}}{16}$$

Now solve the equation x=\frac{-3±\sqrt{1257}}{16} when ± is plus. Add -3 to \sqrt{1257}.

$$x=\frac{\sqrt{1257}-3}{16}$$

Now solve the equation x=\frac{-3±\sqrt{1257}}{16} when ± is minus. Subtract \sqrt{1257} from -3.

$$x=\frac{-\sqrt{1257}-3}{16}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{1257}}{16} for x_{1} and \frac{-3-\sqrt{1257}}{16} for x_{2}.

$$8x^{2}+3x-39=8\left(x-\frac{\sqrt{1257}-3}{16}\right)\left(x-\frac{-\sqrt{1257}-3}{16}\right)$$