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2c^2-3c-135=0
a = 2; b = -3; c = -135;
Δ = b2-4ac
Δ = -32-4·2·(-135)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1089}=33$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-33}{2*2}=\frac{-30}{4} =-7+1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+33}{2*2}=\frac{36}{4} =9 $
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