5x^2+10x=17=0

Solution for 5x^2+10x=17=0 equation:

5x^2+10x=17=0

We move all terms to the left:

5x^2+10x-(17)=0

a = 5; b = 10; c = -17;
Δ = b2-4ac
Δ = 102-4·5·(-17)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{110}}{2*5}=\frac{-10-2\sqrt{110}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{110}}{2*5}=\frac{-10+2\sqrt{110}}{10}
$