4x^2-8x-294=-10

Solution for 4x^2-8x-294=-10 equation:

4x^2-8x-294=-10

We move all terms to the left:

4x^2-8x-294-(-10)=0

We add all the numbers together, and all the variables

4x^2-8x-284=0

a = 4; b = -8; c = -284;
Δ = b2-4ac
Δ = -82-4·4·(-284)
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-48\sqrt{2}}{2*4}=\frac{8-48\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+48\sqrt{2}}{2*4}=\frac{8+48\sqrt{2}}{8}
$