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4x^2+0.4x=15000
We move all terms to the left:
4x^2+0.4x-(15000)=0
a = 4; b = 0.4; c = -15000;
Δ = b2-4ac
Δ = 0.42-4·4·(-15000)
Δ = 240000.16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.4)-\sqrt{240000.16}}{2*4}=\frac{-0.4-\sqrt{240000.16}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{240000.16}}{2*4}=\frac{-0.4+\sqrt{240000.16}}{8} $
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