Solve 5x^2-6x+5+5x^2+8x-8

$$5 x ^ { 2 } - 6 x + 5 + 5 x ^ { 2 } + 8 x - 8$$

Evaluate

$$10x^{2}+2x-3$$

Solution Steps

Combine 5x^{2} and 5x^{2} to get 10x^{2}.

$$10x^{2}-6x+5+8x-8$$

Combine -6x and 8x to get 2x.

$$10x^{2}+2x+5-8$$

Subtract 8 from 5 to get -3.

$$10x^{2}+2x-3$$

Factor

$$10\left(x-\frac{-\sqrt{31}-1}{10}\right)\left(x-\frac{\sqrt{31}-1}{10}\right)$$

Steps Using the Quadratic Formula

Combine 5x^{2} and 5x^{2} to get 10x^{2}.

$$factor(10x^{2}-6x+5+8x-8)$$

Combine -6x and 8x to get 2x.

$$factor(10x^{2}+2x+5-8)$$

Subtract 8 from 5 to get -3.

$$factor(10x^{2}+2x-3)$$

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.

$$10x^{2}+2x-3=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 10\left(-3\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-2±\sqrt{4-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$x=\frac{-2±\sqrt{4+120}}{2\times 10}$$

Add 4 to 120.

$$x=\frac{-2±\sqrt{124}}{2\times 10}$$

Take the square root of 124.

$$x=\frac{-2±2\sqrt{31}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-2±2\sqrt{31}}{20}$$

Now solve the equation x=\frac{-2±2\sqrt{31}}{20} when ± is plus. Add -2 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}-2}{20}$$

Divide -2+2\sqrt{31} by 20.

$$x=\frac{\sqrt{31}-1}{10}$$

Now solve the equation x=\frac{-2±2\sqrt{31}}{20} when ± is minus. Subtract 2\sqrt{31} from -2.

$$x=\frac{-2\sqrt{31}-2}{20}$$

Divide -2-2\sqrt{31} by 20.

$$x=\frac{-\sqrt{31}-1}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{31}}{10} for x_{1} and \frac{-1-\sqrt{31}}{10} for x_{2}.

$$10x^{2}+2x-3=10\left(x-\frac{\sqrt{31}-1}{10}\right)\left(x-\frac{-\sqrt{31}-1}{10}\right)$$