35x^2+4x=35

Solution for 35x^2+4x=35 equation:

35x^2+4x=35

We move all terms to the left:

35x^2+4x-(35)=0

a = 35; b = 4; c = -35;
Δ = b2-4ac
Δ = 42-4·35·(-35)
Δ = 4916
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:

$\sqrt{\Delta}=\sqrt{4916}=\sqrt{4*1229}=\sqrt{4}*\sqrt{1229}=2\sqrt{1229}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{1229}}{2*35}=\frac{-4-2\sqrt{1229}}{70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{1229}}{2*35}=\frac{-4+2\sqrt{1229}}{70}
$