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(2y-48)(2y-4)=180
We move all terms to the left:
(2y-48)(2y-4)-(180)=0
We multiply parentheses ..
(+4y^2-8y-96y+192)-180=0
We get rid of parentheses
4y^2-8y-96y+192-180=0
We add all the numbers together, and all the variables
4y^2-104y+12=0
a = 4; b = -104; c = +12;
Δ = b2-4ac
Δ = -1042-4·4·12
Δ = 10624
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{10624}=\sqrt{64*166}=\sqrt{64}*\sqrt{166}=8\sqrt{166}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-8\sqrt{166}}{2*4}=\frac{104-8\sqrt{166}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+8\sqrt{166}}{2*4}=\frac{104+8\sqrt{166}}{8} $
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