t2+8t=5

Solution for t2+8t=5 equation:

t2+8t=5

We move all terms to the left:

t2+8t-(5)=0

We add all the numbers together, and all the variables

t^2+8t-5=0

a = 1; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·1·(-5)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{21}}{2*1}=\frac{-8-2\sqrt{21}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{21}}{2*1}=\frac{-8+2\sqrt{21}}{2}
$