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0.4x^2+0.5x-0.2=2.4
We move all terms to the left:
0.4x^2+0.5x-0.2-(2.4)=0
We add all the numbers together, and all the variables
0.4x^2+0.5x-2.6=0
a = 0.4; b = 0.5; c = -2.6;
Δ = b2-4ac
Δ = 0.52-4·0.4·(-2.6)
Δ = 4.41
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{-(0.5)-\sqrt{4.41}}{2*0.4}=\frac{-0.5-\sqrt{4.41}}{0.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{4.41}}{2*0.4}=\frac{-0.5+\sqrt{4.41}}{0.8} $
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