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5x^2+15x-140=0
a = 5; b = 15; c = -140;
Δ = b2-4ac
Δ = 152-4·5·(-140)
Δ = 3025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3025}=55$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-55}{2*5}=\frac{-70}{10} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+55}{2*5}=\frac{40}{10} =4 $
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