-3x^2+19x=-7

Solution for -3x^2+19x=-7 equation:

-3x^2+19x=-7

We move all terms to the left:

-3x^2+19x-(-7)=0

We add all the numbers together, and all the variables

-3x^2+19x+7=0

a = -3; b = 19; c = +7;
Δ = b2-4ac
Δ = 192-4·(-3)·7
Δ = 445
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{445}}{2*-3}=\frac{-19-\sqrt{445}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{445}}{2*-3}=\frac{-19+\sqrt{445}}{-6}
$