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3m^2+20m+25=0
a = 3; b = 20; c = +25;
Δ = b2-4ac
Δ = 202-4·3·25
Δ = 100
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{100}=10$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*3}=\frac{-30}{6} =-5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*3}=\frac{-10}{6} =-1+2/3 $
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