-2x^2+14x+38=0

Solution for -2x^2+14x+38=0 equation:

-2x^2+14x+38=0

a = -2; b = 14; c = +38;
Δ = b2-4ac
Δ = 142-4·(-2)·38
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10\sqrt{5}}{2*-2}=\frac{-14-10\sqrt{5}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10\sqrt{5}}{2*-2}=\frac{-14+10\sqrt{5}}{-4}
$