3x^2-40x+64=0

Solution for 3x^2-40x+64=0 equation:

3x^2-40x+64=0

a = 3; b = -40; c = +64;
Δ = b2-4ac
Δ = -402-4·3·64
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{13}}{2*3}=\frac{40-8\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{13}}{2*3}=\frac{40+8\sqrt{13}}{6}
$