4t^2-21t-18=0

Solution for 4t^2-21t-18=0 equation:

4t^2-21t-18=0

a = 4; b = -21; c = -18;
Δ = b2-4ac
Δ = -212-4·4·(-18)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-27}{2*4}=\frac{-6}{8}
=-3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+27}{2*4}=\frac{48}{8}
=6
$