3p^2+40p-675=0

Solution for 3p^2+40p-675=0 equation:

3p^2+40p-675=0

a = 3; b = 40; c = -675;
Δ = b2-4ac
Δ = 402-4·3·(-675)
Δ = 9700
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:

$\sqrt{\Delta}=\sqrt{9700}=\sqrt{100*97}=\sqrt{100}*\sqrt{97}=10\sqrt{97}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{97}}{2*3}=\frac{-40-10\sqrt{97}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{97}}{2*3}=\frac{-40+10\sqrt{97}}{6}
$