t^2+22t-4=0

Solution for t^2+22t-4=0 equation:

t^2+22t-4=0

a = 1; b = 22; c = -4;
Δ = b2-4ac
Δ = 222-4·1·(-4)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-10\sqrt{5}}{2*1}=\frac{-22-10\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+10\sqrt{5}}{2*1}=\frac{-22+10\sqrt{5}}{2}
$