7x^2+23x=196

Solution for 7x^2+23x=196 equation:

7x^2+23x=196

We move all terms to the left:

7x^2+23x-(196)=0

a = 7; b = 23; c = -196;
Δ = b2-4ac
Δ = 232-4·7·(-196)
Δ = 6017
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{6017}}{2*7}=\frac{-23-\sqrt{6017}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{6017}}{2*7}=\frac{-23+\sqrt{6017}}{14}
$