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10x^2+41x-45=0
a = 10; b = 41; c = -45;
Δ = b2-4ac
Δ = 412-4·10·(-45)
Δ = 3481
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{3481}=59$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-59}{2*10}=\frac{-100}{20} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+59}{2*10}=\frac{18}{20} =9/10 $
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