Solve x^2+3x-5

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

Factor

$$\left(x-\frac{-\sqrt{29}-3}{2}\right)\left(x-\frac{\sqrt{29}-3}{2}\right)$$

Steps Using the Quadratic Formula

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.

$$x^{2}+3x-5=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-5\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-3±\sqrt{9+20}}{2}$$

Add 9 to 20.

$$x=\frac{-3±\sqrt{29}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{29}}{2} when ± is plus. Add -3 to \sqrt{29}.

$$x=\frac{\sqrt{29}-3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{29}}{2} when ± is minus. Subtract \sqrt{29} from -3.

$$x=\frac{-\sqrt{29}-3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{29}}{2} for x_{1} and \frac{-3-\sqrt{29}}{2} for x_{2}.

$$x^{2}+3x-5=\left(x-\frac{\sqrt{29}-3}{2}\right)\left(x-\frac{-\sqrt{29}-3}{2}\right)$$

Evaluate

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