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10m^2+16m-42=0
a = 10; b = 16; c = -42;
Δ = b2-4ac
Δ = 162-4·10·(-42)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1936}=44$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-44}{2*10}=\frac{-60}{20} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+44}{2*10}=\frac{28}{20} =1+2/5 $
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