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5x^2+6.4x=100
We move all terms to the left:
5x^2+6.4x-(100)=0
a = 5; b = 6.4; c = -100;
Δ = b2-4ac
Δ = 6.42-4·5·(-100)
Δ = 2040.96
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{-(6.4)-\sqrt{2040.96}}{2*5}=\frac{-6.4-\sqrt{2040.96}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6.4)+\sqrt{2040.96}}{2*5}=\frac{-6.4+\sqrt{2040.96}}{10} $
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