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-0.02x^2+0.12x+0.12=0
a = -0.02; b = 0.12; c = +0.12;
Δ = b2-4ac
Δ = 0.122-4·(-0.02)·0.12
Δ = 0.024
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{-(0.12)-\sqrt{0.024}}{2*-0.02}=\frac{-0.12-\sqrt{0.024}}{-0.04} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.12)+\sqrt{0.024}}{2*-0.02}=\frac{-0.12+\sqrt{0.024}}{-0.04} $
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