-3t^2+26t-40=0

Solution for -3t^2+26t-40=0 equation:

-3t^2+26t-40=0

a = -3; b = 26; c = -40;
Δ = b2-4ac
Δ = 262-4·(-3)·(-40)
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{-(26)-14}{2*-3}=\frac{-40}{-6}
=6+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+14}{2*-3}=\frac{-12}{-6}
=+2
$