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200x+4x^2=2400
We move all terms to the left:
200x+4x^2-(2400)=0
a = 4; b = 200; c = -2400;
Δ = b2-4ac
Δ = 2002-4·4·(-2400)
Δ = 78400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{78400}=280$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-280}{2*4}=\frac{-480}{8} =-60 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+280}{2*4}=\frac{80}{8} =10 $
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