(4w+3)(1-w)=0

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Solution for (4w+3)(1-w)=0 equation:



(4w+3)(1-w)=0
We add all the numbers together, and all the variables
(4w+3)(-1w+1)=0
We multiply parentheses ..
(-4w^2+4w-3w+3)=0
We get rid of parentheses
-4w^2+4w-3w+3=0
We add all the numbers together, and all the variables
-4w^2+w+3=0
a = -4; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-4)·3
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-4}=\frac{-8}{-8} =1 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-4}=\frac{6}{-8} =-3/4 $

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