t2+14t-40=0

Solution for t2+14t-40=0 equation:

t2+14t-40=0

We add all the numbers together, and all the variables

t^2+14t-40=0

a = 1; b = 14; c = -40;
Δ = b2-4ac
Δ = 142-4·1·(-40)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{89}}{2*1}=\frac{-14-2\sqrt{89}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{89}}{2*1}=\frac{-14+2\sqrt{89}}{2}
$