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2.5x^2+7x=3
We move all terms to the left:
2.5x^2+7x-(3)=0
a = 2.5; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·2.5·(-3)
Δ = 79
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{-(7)-\sqrt{79}}{2*2.5}=\frac{-7-\sqrt{79}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{79}}{2*2.5}=\frac{-7+\sqrt{79}}{5} $
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