(2w+3)w=42

Solution for (2w+3)w=42 equation:

(2w+3)w=42

We move all terms to the left:

(2w+3)w-(42)=0

We multiply parentheses

2w^2+3w-42=0

a = 2; b = 3; c = -42;
Δ = b2-4ac
Δ = 32-4·2·(-42)
Δ = 345
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{345}}{2*2}=\frac{-3-\sqrt{345}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{345}}{2*2}=\frac{-3+\sqrt{345}}{4}
$