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4y^2-25=100
We move all terms to the left:
4y^2-25-(100)=0
We add all the numbers together, and all the variables
4y^2-125=0
a = 4; b = 0; c = -125;
Δ = b2-4ac
Δ = 02-4·4·(-125)
Δ = 2000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*4}=\frac{0-20\sqrt{5}}{8} =-\frac{20\sqrt{5}}{8} =-\frac{5\sqrt{5}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*4}=\frac{0+20\sqrt{5}}{8} =\frac{20\sqrt{5}}{8} =\frac{5\sqrt{5}}{2} $
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