What are the solutions to the equation (x+4)(x-3)=-10 ?

The solutions to the equation (x+4)(x-3)=-10 are x=1,x=-2

Find the zeros of (x+4)(x-3)=-10

The zeros of (x+4)(x-3)=-10 are x=1,x=-2

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

x = 1, x = - 2

Solution steps

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

Expand

(x + 4) (x - 3) : x^{2} + x - 12

x^{2} + x - 12 = - 10

Move

10

to the left side

x^{2} + x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

x_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 3}{2 \cdot 1}, x_{2} = \frac{- 1 - 3}{2 \cdot 1}

x = \frac{- 1 + 3}{2 \cdot 1} : 1

x = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 1, x = - 2

6 c^{2} - 24 = 0

25 - x^{2} = 4 x

C, E = M C^{2}

(y - 2)^{2} - 10 (y - 2) = - 9

5 n (n - 4) = 0

What are the solutions to the equation (x+4)(x-3)=-10 ?
The solutions to the equation (x+4)(x-3)=-10 are x=1,x=-2
Find the zeros of (x+4)(x-3)=-10
The zeros of (x+4)(x-3)=-10 are x=1,x=-2