10x2+10x=140

Solution for 10x2+10x=140 equation:

10x^2+10x=140

We move all terms to the left:

10x^2+10x-(140)=0

a = 10; b = 10; c = -140;
Δ = b2-4ac
Δ = 102-4·10·(-140)
Δ = 5700
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:

$\sqrt{\Delta}=\sqrt{5700}=\sqrt{100*57}=\sqrt{100}*\sqrt{57}=10\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{57}}{2*10}=\frac{-10-10\sqrt{57}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{57}}{2*10}=\frac{-10+10\sqrt{57}}{20}
$