20x^2=28x

Solution for 20x^2=28x equation:

20x^2=28x

We move all terms to the left:

20x^2-(28x)=0

a = 20; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·20·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*20}=\frac{0}{40}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*20}=\frac{56}{40}
=1+2/5
$