17x^2+19x=6

Solution for 17x^2+19x=6 equation:

17x^2+19x=6

We move all terms to the left:

17x^2+19x-(6)=0

a = 17; b = 19; c = -6;
Δ = b2-4ac
Δ = 192-4·17·(-6)
Δ = 769
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{-(19)-\sqrt{769}}{2*17}=\frac{-19-\sqrt{769}}{34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{769}}{2*17}=\frac{-19+\sqrt{769}}{34}
$