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10b^2+22b-72=0
a = 10; b = 22; c = -72;
Δ = b2-4ac
Δ = 222-4·10·(-72)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3364}=58$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-58}{2*10}=\frac{-80}{20} =-4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+58}{2*10}=\frac{36}{20} =1+4/5 $
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