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50+10(2w^2)=2050
We move all terms to the left:
50+10(2w^2)-(2050)=0
We add all the numbers together, and all the variables
102w^2-2000=0
a = 102; b = 0; c = -2000;
Δ = b2-4ac
Δ = 02-4·102·(-2000)
Δ = 816000
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{816000}=\sqrt{1600*510}=\sqrt{1600}*\sqrt{510}=40\sqrt{510}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{510}}{2*102}=\frac{0-40\sqrt{510}}{204} =-\frac{40\sqrt{510}}{204} =-\frac{10\sqrt{510}}{51} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{510}}{2*102}=\frac{0+40\sqrt{510}}{204} =\frac{40\sqrt{510}}{204} =\frac{10\sqrt{510}}{51} $
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