t^2+14t+32=0

Solution for t^2+14t+32=0 equation:

t^2+14t+32=0

a = 1; b = 14; c = +32;
Δ = b2-4ac
Δ = 142-4·1·32
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{17}}{2*1}=\frac{-14-2\sqrt{17}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{17}}{2*1}=\frac{-14+2\sqrt{17}}{2}
$