t^2-5t+14^=0

Solution for t^2-5t+14^=0 equation:

t^2-5t+14^=0

We add all the numbers together, and all the variables

t^2-5t=0

a = 1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·1·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*1}=\frac{0}{2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*1}=\frac{10}{2}
=5
$