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4(x+2)(x+5)=280^3
We move all terms to the left:
4(x+2)(x+5)-(280^3)=0
We add all the numbers together, and all the variables
4(x+2)(x+5)-21952000=0
We multiply parentheses ..
4(+x^2+5x+2x+10)-21952000=0
We multiply parentheses
4x^2+20x+8x+40-21952000=0
We add all the numbers together, and all the variables
4x^2+28x-21951960=0
a = 4; b = 28; c = -21951960;
Δ = b2-4ac
Δ = 282-4·4·(-21951960)
Δ = 351232144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{351232144}=\sqrt{16*21952009}=\sqrt{16}*\sqrt{21952009}=4\sqrt{21952009}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{21952009}}{2*4}=\frac{-28-4\sqrt{21952009}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{21952009}}{2*4}=\frac{-28+4\sqrt{21952009}}{8} $
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