Question

(u + 3)(u - 4) = -10

Answer 1

(u + 3) ( u - 4) = - 10
u^2 + 3u - 4u - 12 + 10 = 0
u^2 - u - 2 = 0
D = 1 - 4 (- 2)* 1 = 1 + 8 = 9 ;  YD = 3
u1 = (1 + 3) / 2 = 2
u2 = (1 - 3) /2 = - 1

Answer 2

U^2-4u+3u-12+10=0
U^2-1u-2=0
D=1-4*1*(-2)=1+8=9=3^2
X1=(1+3)/2=2
X2=(1-3)/2=-1

Answer 3

To solve the equation $(u + 3)(u - 4) = -10$, we'll follow these steps:

1. Expand the left-hand side of the equation.
2. Simplify the expression.
3. Solve for $u$.

Let's go through each step:

Step 1: Expand the left-hand side. $(u + 3)(u - 4) = u^2 - 4u + 3u - 12$

Step 2: Simplify the expression. $u^2 - u - 12 = -10$

Step 3: Bring all terms to one side to set the equation to zero. $u^2 - u - 12 + 10 = 0$

$u^2 - u - 2 = 0$

This is now a quadratic equation in the form $au^2 + bu + c = 0$ where $a = 1$, $b = -1$, and $c = -2$.

To solve for $u$, we can use the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in the values $a = 1$, $b = -1$, and $c = -2$, we get:

$u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

$u = \frac{1 \pm \sqrt{1 + 8}}{2}$

$u = \frac{1 \pm \sqrt{9}}{2}$

$u = \frac{1 \pm 3}{2}$

This gives two possible values for $u$:

1. $u = \frac{1 + 3}{2} = 2$
2. $u = \frac{1 - 3}{2} = -1$

So, the solutions to the equation $(u + 3)(u - 4) = -10$ are $u = 2$ and $u = -1$.