t^2+4t-7=0

Solution for t^2+4t-7=0 equation:

t^2+4t-7=0

a = 1; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·1·(-7)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{11}}{2*1}=\frac{-4-2\sqrt{11}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{11}}{2*1}=\frac{-4+2\sqrt{11}}{2}
$