t^2+26t-17=0

Solution for t^2+26t-17=0 equation:

t^2+26t-17=0

a = 1; b = 26; c = -17;
Δ = b2-4ac
Δ = 262-4·1·(-17)
Δ = 744
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:

$\sqrt{\Delta}=\sqrt{744}=\sqrt{4*186}=\sqrt{4}*\sqrt{186}=2\sqrt{186}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{186}}{2*1}=\frac{-26-2\sqrt{186}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{186}}{2*1}=\frac{-26+2\sqrt{186}}{2}
$