Solve 3x^2+10x-5

$$3 { x }^{ 2 } +10x-5$$

Factor

$$3\left(x-\frac{-2\sqrt{10}-5}{3}\right)\left(x-\frac{2\sqrt{10}-5}{3}\right)$$

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.

$$3x^{2}+10x-5=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-10±\sqrt{100-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$x=\frac{-10±\sqrt{100+60}}{2\times 3}$$

Add 100 to 60.

$$x=\frac{-10±\sqrt{160}}{2\times 3}$$

Take the square root of 160.

$$x=\frac{-10±4\sqrt{10}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-10±4\sqrt{10}}{6}$$

Now solve the equation x=\frac{-10±4\sqrt{10}}{6} when ± is plus. Add -10 to 4\sqrt{10}.

$$x=\frac{4\sqrt{10}-10}{6}$$

Divide -10+4\sqrt{10} by 6.

$$x=\frac{2\sqrt{10}-5}{3}$$

Now solve the equation x=\frac{-10±4\sqrt{10}}{6} when ± is minus. Subtract 4\sqrt{10} from -10.

$$x=\frac{-4\sqrt{10}-10}{6}$$

Divide -10-4\sqrt{10} by 6.

$$x=\frac{-2\sqrt{10}-5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+2\sqrt{10}}{3} for x_{1} and \frac{-5-2\sqrt{10}}{3} for x_{2}.

$$3x^{2}+10x-5=3\left(x-\frac{2\sqrt{10}-5}{3}\right)\left(x-\frac{-2\sqrt{10}-5}{3}\right)$$

Evaluate

$$3x^{2}+10x-5$$