7x^2+19x=80

Solution for 7x^2+19x=80 equation:

7x^2+19x=80

We move all terms to the left:

7x^2+19x-(80)=0

a = 7; b = 19; c = -80;
Δ = b2-4ac
Δ = 192-4·7·(-80)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-51}{2*7}=\frac{-70}{14}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+51}{2*7}=\frac{32}{14}
=2+2/7
$