-45x^2+2x+10=0

Solution for -45x^2+2x+10=0 equation:

-45x^2+2x+10=0

a = -45; b = 2; c = +10;
Δ = b2-4ac
Δ = 22-4·(-45)·10
Δ = 1804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1804}=\sqrt{4*451}=\sqrt{4}*\sqrt{451}=2\sqrt{451}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{451}}{2*-45}=\frac{-2-2\sqrt{451}}{-90}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{451}}{2*-45}=\frac{-2+2\sqrt{451}}{-90}
$