t^2+14t-3=0

Solution for t^2+14t-3=0 equation:

t^2+14t-3=0

a = 1; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·1·(-3)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{13}}{2*1}=\frac{-14-4\sqrt{13}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{13}}{2*1}=\frac{-14+4\sqrt{13}}{2}
$