Answer :

konrad509
[tex] 4b^2+8b+7=4\\ 4b^2+8b+3=0\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{-8\pm\sqrt{8^2-4\cdot4\cdot3}}{2\cdot4}\\ x=\dfrac{-8\pm\sqrt{64-48}}{8}\\ x=\dfrac{-8\pm\sqrt{16}}{8}\\ x=\dfrac{-8\pm4}{8}\\ x=\dfrac{-2\pm1}{2}\\ x=\dfrac{-1}{2 }\vee x=\dfrac{-3}{2}\\ x=-\dfrac{1}{2 }\vee x=-\dfrac{3}{2}[/tex]
TalhaHasan
b= { (-8) ±    Sqrt(8^2)-4(4x(3)  )  } / 2(4)
b={ (-8±)  (Sqrt(16) )  } / 8
b=(-8+4)/8 and b=(-8-4)/8
b= -4/8    and b=-12/8
b=-1/2  and b=-3/2
b={-1/2, -3/2} is the solution.
Click image for formula.
Sqrt means Square Root.

View image TalhaHasan

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