-4x^2+20x+49=0

Solution for -4x^2+20x+49=0 equation:

-4x^2+20x+49=0

a = -4; b = 20; c = +49;
Δ = b2-4ac
Δ = 202-4·(-4)·49
Δ = 1184
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:

$\sqrt{\Delta}=\sqrt{1184}=\sqrt{16*74}=\sqrt{16}*\sqrt{74}=4\sqrt{74}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{74}}{2*-4}=\frac{-20-4\sqrt{74}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{74}}{2*-4}=\frac{-20+4\sqrt{74}}{-8}
$