42x^2+80x+32=0

Solution for 42x^2+80x+32=0 equation:

42x^2+80x+32=0

a = 42; b = 80; c = +32;
Δ = b2-4ac
Δ = 802-4·42·32
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-32}{2*42}=\frac{-112}{84}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+32}{2*42}=\frac{-48}{84}
=-4/7
$