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2.25x^2+2.8125x-560=0
a = 2.25; b = 2.8125; c = -560;
Δ = b2-4ac
Δ = 2.81252-4·2.25·(-560)
Δ = 5047.91015625
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{-(2.8125)-\sqrt{5047.91015625}}{2*2.25}=\frac{-2.8125-\sqrt{5047.91015625}}{4.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.8125)+\sqrt{5047.91015625}}{2*2.25}=\frac{-2.8125+\sqrt{5047.91015625}}{4.5} $
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