t^2-4t=18

Solution for t^2-4t=18 equation:

t^2-4t=18

We move all terms to the left:

t^2-4t-(18)=0

a = 1; b = -4; c = -18;
Δ = b2-4ac
Δ = -42-4·1·(-18)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{22}}{2*1}=\frac{4-2\sqrt{22}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{22}}{2*1}=\frac{4+2\sqrt{22}}{2}
$