t^2+8t-21=0

Solution for t^2+8t-21=0 equation:

t^2+8t-21=0

a = 1; b = 8; c = -21;
Δ = b2-4ac
Δ = 82-4·1·(-21)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{37}}{2*1}=\frac{-8-2\sqrt{37}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{37}}{2*1}=\frac{-8+2\sqrt{37}}{2}
$