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2x^2+15x-203=0
a = 2; b = 15; c = -203;
Δ = b2-4ac
Δ = 152-4·2·(-203)
Δ = 1849
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{1849}=43$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-43}{2*2}=\frac{-58}{4} =-14+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+43}{2*2}=\frac{28}{4} =7 $
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