3t^2+27t+42=0

Solution for 3t^2+27t+42=0 equation:

3t^2+27t+42=0

a = 3; b = 27; c = +42;
Δ = b2-4ac
Δ = 272-4·3·42
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-15}{2*3}=\frac{-42}{6}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+15}{2*3}=\frac{-12}{6}
=-2
$