4x^2+23x=35

Solution for 4x^2+23x=35 equation:

4x^2+23x=35

We move all terms to the left:

4x^2+23x-(35)=0

a = 4; b = 23; c = -35;
Δ = b2-4ac
Δ = 232-4·4·(-35)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*4}=\frac{-56}{8}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*4}=\frac{10}{8}
=1+1/4
$