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-0.45x^2+2.4x+3=0
a = -0.45; b = 2.4; c = +3;
Δ = b2-4ac
Δ = 2.42-4·(-0.45)·3
Δ = 11.16
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{-(2.4)-\sqrt{11.16}}{2*-0.45}=\frac{-2.4-\sqrt{11.16}}{-0.9} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.4)+\sqrt{11.16}}{2*-0.45}=\frac{-2.4+\sqrt{11.16}}{-0.9} $
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