4p^2+13p=35

Solution for 4p^2+13p=35 equation:

4p^2+13p=35

We move all terms to the left:

4p^2+13p-(35)=0

a = 4; b = 13; c = -35;
Δ = b2-4ac
Δ = 132-4·4·(-35)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-27}{2*4}=\frac{-40}{8}
=-5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+27}{2*4}=\frac{14}{8}
=1+3/4
$