x^2+70x=544=0

Solution for x^2+70x=544=0 equation:

x^2+70x=544=0

We move all terms to the left:

x^2+70x-(544)=0

a = 1; b = 70; c = -544;
Δ = b2-4ac
Δ = 702-4·1·(-544)
Δ = 7076
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:

$\sqrt{\Delta}=\sqrt{7076}=\sqrt{4*1769}=\sqrt{4}*\sqrt{1769}=2\sqrt{1769}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1769}}{2*1}=\frac{-70-2\sqrt{1769}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1769}}{2*1}=\frac{-70+2\sqrt{1769}}{2}
$