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25x^2+42+65x=0
a = 25; b = 65; c = +42;
Δ = b2-4ac
Δ = 652-4·25·42
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(65)-5}{2*25}=\frac{-70}{50} =-1+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+5}{2*25}=\frac{-60}{50} =-1+1/5 $
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