Find the roots of the following quadratic equ | vedclass

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(B) Given equation: $(x+4)(x+5)=3(x+1)(x+2)+2x$Expanding both sides:$x^2 + 9x + 20 = 3(x^2 + 3x + 2) + 2x$$x^2 + 9x + 20 = 3x^2 + 9x + 6 + 2x$Rearrang

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$\frac{-1+\sqrt{29}}{2}$ and $\frac{-1-\sqrt{29}}{2}$

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(B) Given equation: $(x+4)(x+5)=3(x+1)(x+2)+2x$Expanding both sides:$x^2 + 9x + 20 = 3(x^2 + 3x + 2) + 2x$$x^2 + 9x + 20 = 3x^2 + 9x + 6 + 2x$Rearranging terms to form $ax^2 + bx + c = 0$:$x^2 - 3x^2 + 9x - 9x - 2x + 20 - 6 = 0$$-2x^2 - 2x + 14 = 0$Dividing by $-2$:$x^2 + x - 7 = 0$Comparing with $ax^2 + bx + c = 0$,we get $a=1, b=1, c=-7$.Discriminant $D = b^2 - 4ac = (1)^2 - 4(1)(-7) = 1 + 28 = 29$.Since $D > 0$,real and distinct roots exist.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{-1 \pm \sqrt{29}}{2(1)}$$x = \frac{-1 \pm \sqrt{29}}{2}$Thus,the roots are $\frac{-1+\sqrt{29}}{2}$ and $\frac{-1-\sqrt{29}}{2}$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $(x+4)(x+5)=3(x+1)(x+2)+2x$

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