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49x^2+21x-10=0
a = 49; b = 21; c = -10;
Δ = b2-4ac
Δ = 212-4·49·(-10)
Δ = 2401
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{2401}=49$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-49}{2*49}=\frac{-70}{98} =-5/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+49}{2*49}=\frac{28}{98} =2/7 $
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