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2x^2-3x-189=0
a = 2; b = -3; c = -189;
Δ = b2-4ac
Δ = -32-4·2·(-189)
Δ = 1521
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{1521}=39$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-39}{2*2}=\frac{-36}{4} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+39}{2*2}=\frac{42}{4} =10+1/2 $
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