$$5x^{2}-40x+5=0$$
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.$$x=\sqrt{15}+4$$ $$x=4-\sqrt{15}$$