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2.4x^2+3x-22=0
a = 2.4; b = 3; c = -22;
Δ = b2-4ac
Δ = 32-4·2.4·(-22)
Δ = 220.2
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)-\sqrt{220.2}}{2*2.4}=\frac{-3-\sqrt{220.2}}{4.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{220.2}}{2*2.4}=\frac{-3+\sqrt{220.2}}{4.8} $
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