Question

2) (x-8)(2x+3)=(3x-5)(x+4);​

Answer 1

(х-8)(2х+3)=(3х-5)(х+4)

2х²+3х-16х-24=3х²+12х-5х-20

2х²-13х-24-3х²-7х+20=0

-х²-20х-4=0

х²+20х+4=0

D=b²-4ac=20²-4×1×4=400-16=384

x₁=(-20-√384)/2

x₂=(-20+√384)/2

x₁=(-20-8√6)/2

x₂=(-20+8√6)/2

x₁=2(-10-4√6)/2

x₂=2(-10+4√6)/2

x₁=-10-4√6

x₂=-10+4√6

______________

x₁=-10-4√6; x₂=4√6-10.

Answer 2

To solve the given equation $(x-8)(2x+3) = (3x-5)(x+4)$, we'll start by expanding both sides of the equation and then simplify to solve for $x$.

Given equation: $(x-8)(2x+3) = (3x-5)(x+4)$

Expanding both sides:

Left side: $2x^2 + 3x - 16x - 24$ $2x^2 - 13x - 24$

Right side: $3x^2 + 4x - 5x - 20$ $3x^2 - x - 20$

So, the equation becomes: $2x^2 - 13x - 24 = 3x^2 - x - 20$

Now, let's solve for $x$:

Subtract $2x^2$ from both sides: $-13x - 24 = x^2 - x - 20$

Subtract $x^2 - x - 20$ from both sides: $-12x - 4 = 0$

Divide both sides by -4: $3x + 1 = 0$

Subtract 1 from both sides: $3x = -1$

Divide by 3: $x = -\frac{1}{3}$

So, the solution for the equation is $x = -\frac{1}{3}$.