(18a^2+51a-42)=0

Solution for (18a^2+51a-42)=0 equation:

(18a^2+51a-42)=0

We get rid of parentheses

18a^2+51a-42=0

a = 18; b = 51; c = -42;
Δ = b2-4ac
Δ = 512-4·18·(-42)
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-75}{2*18}=\frac{-126}{36}
=-3+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+75}{2*18}=\frac{24}{36}
=2/3
$