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10x(x+4)=99
We move all terms to the left:
10x(x+4)-(99)=0
We multiply parentheses
10x^2+40x-99=0
a = 10; b = 40; c = -99;
Δ = b2-4ac
Δ = 402-4·10·(-99)
Δ = 5560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5560}=\sqrt{4*1390}=\sqrt{4}*\sqrt{1390}=2\sqrt{1390}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{1390}}{2*10}=\frac{-40-2\sqrt{1390}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{1390}}{2*10}=\frac{-40+2\sqrt{1390}}{20} $
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