P-3q2+10q=5

Solution for P-3q2+10q=5 equation:

-3P^2+10P=5

We move all terms to the left:

-3P^2+10P-(5)=0

a = -3; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·(-3)·(-5)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{10}}{2*-3}=\frac{-10-2\sqrt{10}}{-6}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{10}}{2*-3}=\frac{-10+2\sqrt{10}}{-6}
$