-5x^2+2x+35=0

Solution for -5x^2+2x+35=0 equation:

-5x^2+2x+35=0

a = -5; b = 2; c = +35;
Δ = b2-4ac
Δ = 22-4·(-5)·35
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8\sqrt{11}}{2*-5}=\frac{-2-8\sqrt{11}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8\sqrt{11}}{2*-5}=\frac{-2+8\sqrt{11}}{-10}
$