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-2.4q^2+2.8q+495=0
a = -2.4; b = 2.8; c = +495;
Δ = b2-4ac
Δ = 2.82-4·(-2.4)·495
Δ = 4759.84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.8)-\sqrt{4759.84}}{2*-2.4}=\frac{-2.8-\sqrt{4759.84}}{-4.8} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.8)+\sqrt{4759.84}}{2*-2.4}=\frac{-2.8+\sqrt{4759.84}}{-4.8} $
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