7w^2+12w+10=4w^2

Solution for 7w^2+12w+10=4w^2 equation:

7w^2+12w+10=4w^2

We move all terms to the left:

7w^2+12w+10-(4w^2)=0

determiningTheFunctionDomain
7w^2-4w^2+12w+10=0

We add all the numbers together, and all the variables

3w^2+12w+10=0

a = 3; b = 12; c = +10;
Δ = b2-4ac
Δ = 122-4·3·10
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{6}}{2*3}=\frac{-12-2\sqrt{6}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{6}}{2*3}=\frac{-12+2\sqrt{6}}{6}
$