-2t^2-32t=42

Solution for -2t^2-32t=42 equation:

-2t^2-32t=42

We move all terms to the left:

-2t^2-32t-(42)=0

a = -2; b = -32; c = -42;
Δ = b2-4ac
Δ = -322-4·(-2)·(-42)
Δ = 688
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{43}}{2*-2}=\frac{32-4\sqrt{43}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{43}}{2*-2}=\frac{32+4\sqrt{43}}{-4}
$