12x^2+28x-17=0

Solution for 12x^2+28x-17=0 equation:

12x^2+28x-17=0

a = 12; b = 28; c = -17;
Δ = b2-4ac
Δ = 282-4·12·(-17)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-40}{2*12}=\frac{-68}{24}
=-2+5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+40}{2*12}=\frac{12}{24}
=1/2
$