Solve for: 5 x ^ 2-40 x+5 = 0

Expression: $$5 x ^ { 2 } - 40 x + 5 = 0$$

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.

$$5x^{2}-40x+5=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-40$ for $b$, and $5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Square $-40$.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\times 5}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\times 5}}{2\times 5}$$

Multiply $-20$ times $5$.

$$x=\frac{-\left(-40\right)±\sqrt{1600-100}}{2\times 5}$$

Add $1600$ to $-100$.

$$x=\frac{-\left(-40\right)±\sqrt{1500}}{2\times 5}$$

Take the square root of $1500$.

$$x=\frac{-\left(-40\right)±10\sqrt{15}}{2\times 5}$$

The opposite of $-40$ is $40$.

$$x=\frac{40±10\sqrt{15}}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{40±10\sqrt{15}}{10}$$

Now solve the equation $x=\frac{40±10\sqrt{15}}{10}$ when $±$ is plus. Add $40$ to $10\sqrt{15}$.

$$x=\frac{10\sqrt{15}+40}{10}$$

Divide $40+10\sqrt{15}$ by $10$.

$$x=\sqrt{15}+4$$

Now solve the equation $x=\frac{40±10\sqrt{15}}{10}$ when $±$ is minus. Subtract $10\sqrt{15}$ from $40$.

$$x=\frac{40-10\sqrt{15}}{10}$$

Divide $40-10\sqrt{15}$ by $10$.

$$x=4-\sqrt{15}$$

The equation is now solved.

$$x=\sqrt{15}+4$$ $$x=4-\sqrt{15}$$