19x^2+10x=90

Solution for 19x^2+10x=90 equation:

19x^2+10x=90

We move all terms to the left:

19x^2+10x-(90)=0

a = 19; b = 10; c = -90;
Δ = b2-4ac
Δ = 102-4·19·(-90)
Δ = 6940
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:

$\sqrt{\Delta}=\sqrt{6940}=\sqrt{4*1735}=\sqrt{4}*\sqrt{1735}=2\sqrt{1735}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1735}}{2*19}=\frac{-10-2\sqrt{1735}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1735}}{2*19}=\frac{-10+2\sqrt{1735}}{38}
$