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2x(2x+10)=420
We move all terms to the left:
2x(2x+10)-(420)=0
We multiply parentheses
4x^2+20x-420=0
a = 4; b = 20; c = -420;
Δ = b2-4ac
Δ = 202-4·4·(-420)
Δ = 7120
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7120}=\sqrt{16*445}=\sqrt{16}*\sqrt{445}=4\sqrt{445}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{445}}{2*4}=\frac{-20-4\sqrt{445}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{445}}{2*4}=\frac{-20+4\sqrt{445}}{8} $
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