t^2+2t=47

Solution for t^2+2t=47 equation:

t^2+2t=47

We move all terms to the left:

t^2+2t-(47)=0

a = 1; b = 2; c = -47;
Δ = b2-4ac
Δ = 22-4·1·(-47)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8\sqrt{3}}{2*1}=\frac{-2-8\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8\sqrt{3}}{2*1}=\frac{-2+8\sqrt{3}}{2}
$