35x^2+20=53x

Solution for 35x^2+20=53x equation:

35x^2+20=53x

We move all terms to the left:

35x^2+20-(53x)=0

a = 35; b = -53; c = +20;
Δ = b2-4ac
Δ = -532-4·35·20
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-3}{2*35}=\frac{50}{70}
=5/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+3}{2*35}=\frac{56}{70}
=4/5
$