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(4y+2)(y+5)=280
We move all terms to the left:
(4y+2)(y+5)-(280)=0
We multiply parentheses ..
(+4y^2+20y+2y+10)-280=0
We get rid of parentheses
4y^2+20y+2y+10-280=0
We add all the numbers together, and all the variables
4y^2+22y-270=0
a = 4; b = 22; c = -270;
Δ = b2-4ac
Δ = 222-4·4·(-270)
Δ = 4804
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4804}=\sqrt{4*1201}=\sqrt{4}*\sqrt{1201}=2\sqrt{1201}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{1201}}{2*4}=\frac{-22-2\sqrt{1201}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{1201}}{2*4}=\frac{-22+2\sqrt{1201}}{8} $
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