0=-3x^2+28x-64

Solution for 0=-3x^2+28x-64 equation:

0=-3x^2+28x-64

We move all terms to the left:

0-(-3x^2+28x-64)=0

We add all the numbers together, and all the variables

-(-3x^2+28x-64)=0

We get rid of parentheses

3x^2-28x+64=0

a = 3; b = -28; c = +64;
Δ = b2-4ac
Δ = -282-4·3·64
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*3}=\frac{24}{6}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*3}=\frac{32}{6}
=5+1/3
$