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2a^2+a-45=0
a = 2; b = 1; c = -45;
Δ = b2-4ac
Δ = 12-4·2·(-45)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*2}=\frac{-20}{4} =-5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*2}=\frac{18}{4} =4+1/2 $
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