Solve 2x^2+3x-5-3

$$2 { x }^{ 2 } +3x-5-3$$

Factor

$$2\left(x-\frac{-\sqrt{73}-3}{4}\right)\left(x-\frac{\sqrt{73}-3}{4}\right)$$

Steps Using the Quadratic Formula

Subtract 3 from -5 to get -8.

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

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.

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

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.

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

Square 3.

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

Multiply -4 times 2.

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

Multiply -8 times -8.

$$x=\frac{-3±\sqrt{9+64}}{2\times 2}$$

Add 9 to 64.

$$x=\frac{-3±\sqrt{73}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±\sqrt{73}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{73}}{4} when ± is plus. Add -3 to \sqrt{73}.

$$x=\frac{\sqrt{73}-3}{4}$$

Now solve the equation x=\frac{-3±\sqrt{73}}{4} when ± is minus. Subtract \sqrt{73} from -3.

$$x=\frac{-\sqrt{73}-3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{73}}{4} for x_{1} and \frac{-3-\sqrt{73}}{4} for x_{2}.

$$2x^{2}+3x-8=2\left(x-\frac{\sqrt{73}-3}{4}\right)\left(x-\frac{-\sqrt{73}-3}{4}\right)$$

Evaluate

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

Solution Steps

Subtract 3 from -5 to get -8.

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