t2+4t-25=0

Solution for t2+4t-25=0 equation:

t2+4t-25=0

We add all the numbers together, and all the variables

t^2+4t-25=0

a = 1; b = 4; c = -25;
Δ = b2-4ac
Δ = 42-4·1·(-25)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{29}}{2*1}=\frac{-4-2\sqrt{29}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{29}}{2*1}=\frac{-4+2\sqrt{29}}{2}
$