5x^2+2x-8x+10=27

Solution for 5x^2+2x-8x+10=27 equation:

5x^2+2x-8x+10=27

We move all terms to the left:

5x^2+2x-8x+10-(27)=0

We add all the numbers together, and all the variables

5x^2-6x-17=0

a = 5; b = -6; c = -17;
Δ = b2-4ac
Δ = -62-4·5·(-17)
Δ = 376
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:

$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{94}}{2*5}=\frac{6-2\sqrt{94}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{94}}{2*5}=\frac{6+2\sqrt{94}}{10}
$