-4(2w-5);w=-3

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Solution for -4(2w-5);w=-3 equation:



-4(2w-5)w=-3
We move all terms to the left:
-4(2w-5)w-(-3)=0
We add all the numbers together, and all the variables
-4(2w-5)w+3=0
We multiply parentheses
-8w^2+20w+3=0
a = -8; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·(-8)·3
Δ = 496
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:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{31}}{2*-8}=\frac{-20-4\sqrt{31}}{-16} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{31}}{2*-8}=\frac{-20+4\sqrt{31}}{-16} $

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