t2+4t-45=0

Solution for t2+4t-45=0 equation:

t2+4t-45=0

We add all the numbers together, and all the variables

t^2+4t-45=0

a = 1; b = 4; c = -45;
Δ = b2-4ac
Δ = 42-4·1·(-45)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*1}=\frac{-18}{2}
=-9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*1}=\frac{10}{2}
=5
$