3p^2+10p-13=4

Solution for 3p^2+10p-13=4 equation:

3p^2+10p-13=4

We move all terms to the left:

3p^2+10p-13-(4)=0

We add all the numbers together, and all the variables

3p^2+10p-17=0

a = 3; b = 10; c = -17;
Δ = b2-4ac
Δ = 102-4·3·(-17)
Δ = 304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{19}}{2*3}=\frac{-10-4\sqrt{19}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{19}}{2*3}=\frac{-10+4\sqrt{19}}{6}
$