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X2+3.5X+0.3=0
We add all the numbers together, and all the variables
X^2+3.5X+0.3=0
a = 1; b = 3.5; c = +0.3;
Δ = b2-4ac
Δ = 3.52-4·1·0.3
Δ = 11.05
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{-(3.5)-\sqrt{11.05}}{2*1}=\frac{-3.5-\sqrt{11.05}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{11.05}}{2*1}=\frac{-3.5+\sqrt{11.05}}{2} $
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