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7x(3x+1)=560
We move all terms to the left:
7x(3x+1)-(560)=0
We multiply parentheses
21x^2+7x-560=0
a = 21; b = 7; c = -560;
Δ = b2-4ac
Δ = 72-4·21·(-560)
Δ = 47089
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{47089}=217$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-217}{2*21}=\frac{-224}{42} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+217}{2*21}=\frac{210}{42} =5 $
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