12x^2+16x-35=0

Solution for 12x^2+16x-35=0 equation:

12x^2+16x-35=0

a = 12; b = 16; c = -35;
Δ = b2-4ac
Δ = 162-4·12·(-35)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-44}{2*12}=\frac{-60}{24}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+44}{2*12}=\frac{28}{24}
=1+1/6
$