16x^2+42x-135=0

Solution for 16x^2+42x-135=0 equation:

16x^2+42x-135=0

a = 16; b = 42; c = -135;
Δ = b2-4ac
Δ = 422-4·16·(-135)
Δ = 10404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10404}=102$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-102}{2*16}=\frac{-144}{32}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+102}{2*16}=\frac{60}{32}
=1+7/8
$