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2a^2+15a+25=0
a = 2; b = 15; c = +25;
Δ = b2-4ac
Δ = 152-4·2·25
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*2}=\frac{-20}{4} =-5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*2}=\frac{-10}{4} =-2+1/2 $
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