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(-12w)(-12w)+20w-3=0
We add all the numbers together, and all the variables
20w+(-12w)(-12w)-3=0
We multiply parentheses ..
(+144w^2)+20w-3=0
We get rid of parentheses
144w^2+20w-3=0
a = 144; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·144·(-3)
Δ = 2128
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2128}=\sqrt{16*133}=\sqrt{16}*\sqrt{133}=4\sqrt{133}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{133}}{2*144}=\frac{-20-4\sqrt{133}}{288} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{133}}{2*144}=\frac{-20+4\sqrt{133}}{288} $
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