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4m^2+5m-21=0
a = 4; b = 5; c = -21;
Δ = b2-4ac
Δ = 52-4·4·(-21)
Δ = 361
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{361}=19$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-19}{2*4}=\frac{-24}{8} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+19}{2*4}=\frac{14}{8} =1+3/4 $
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