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(3y+53)(7y-53)=180
We move all terms to the left:
(3y+53)(7y-53)-(180)=0
We multiply parentheses ..
(+21y^2-159y+371y-2809)-180=0
We get rid of parentheses
21y^2-159y+371y-2809-180=0
We add all the numbers together, and all the variables
21y^2+212y-2989=0
a = 21; b = 212; c = -2989;
Δ = b2-4ac
Δ = 2122-4·21·(-2989)
Δ = 296020
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{296020}=\sqrt{1444*205}=\sqrt{1444}*\sqrt{205}=38\sqrt{205}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(212)-38\sqrt{205}}{2*21}=\frac{-212-38\sqrt{205}}{42} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(212)+38\sqrt{205}}{2*21}=\frac{-212+38\sqrt{205}}{42} $
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