-9p^2+10=0

Solution for -9p^2+10=0 equation:

-9p^2+10=0

a = -9; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-9)·10
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*-9}=\frac{0-6\sqrt{10}}{-18}
=-\frac{6\sqrt{10}}{-18}
=-\frac{\sqrt{10}}{-3}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*-9}=\frac{0+6\sqrt{10}}{-18}
=\frac{6\sqrt{10}}{-18}
=\frac{\sqrt{10}}{-3}
$