Solve (2x^2+3x-1)+(6x^2-3)=

$$( 2 x ^ { 2 } + 3 x - 1 ) + ( 6 x ^ { 2 } - 3 ) =$$

Evaluate

$$8x^{2}+3x-4$$

Solution Steps

Combine 2x^{2} and 6x^{2} to get 8x^{2}.

$$8x^{2}+3x-1-3$$

Subtract 3 from -1 to get -4.

$$8x^{2}+3x-4$$

Factor

$$8\left(x-\frac{-\sqrt{137}-3}{16}\right)\left(x-\frac{\sqrt{137}-3}{16}\right)$$

Steps Using the Quadratic Formula

Combine 2x^{2} and 6x^{2} to get 8x^{2}.

$$factor(8x^{2}+3x-1-3)$$

Subtract 3 from -1 to get -4.

$$factor(8x^{2}+3x-4)$$

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.

$$8x^{2}+3x-4=0$$

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.

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

Square 3.

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

Multiply -4 times 8.

$$x=\frac{-3±\sqrt{9-32\left(-4\right)}}{2\times 8}$$

Multiply -32 times -4.

$$x=\frac{-3±\sqrt{9+128}}{2\times 8}$$

Add 9 to 128.

$$x=\frac{-3±\sqrt{137}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-3±\sqrt{137}}{16}$$

Now solve the equation x=\frac{-3±\sqrt{137}}{16} when ± is plus. Add -3 to \sqrt{137}.

$$x=\frac{\sqrt{137}-3}{16}$$

Now solve the equation x=\frac{-3±\sqrt{137}}{16} when ± is minus. Subtract \sqrt{137} from -3.

$$x=\frac{-\sqrt{137}-3}{16}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{137}}{16} for x_{1} and \frac{-3-\sqrt{137}}{16} for x_{2}.

$$8x^{2}+3x-4=8\left(x-\frac{\sqrt{137}-3}{16}\right)\left(x-\frac{-\sqrt{137}-3}{16}\right)$$