20q2-90+40=0

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Solution for 20q2-90+40=0 equation:



20q^2-90+40=0
We add all the numbers together, and all the variables
20q^2-50=0
a = 20; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·20·(-50)
Δ = 4000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{4000}=\sqrt{400*10}=\sqrt{400}*\sqrt{10}=20\sqrt{10}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{10}}{2*20}=\frac{0-20\sqrt{10}}{40} =-\frac{20\sqrt{10}}{40} =-\frac{\sqrt{10}}{2} $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{10}}{2*20}=\frac{0+20\sqrt{10}}{40} =\frac{20\sqrt{10}}{40} =\frac{\sqrt{10}}{2} $

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