What are the solutions to the equation (m+4)(m-5)=-8 ?

The solutions to the equation (m+4)(m-5)=-8 are m=4,m=-3

Find the zeros of (m+4)(m-5)=-8

The zeros of (m+4)(m-5)=-8 are m=4,m=-3

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(m + 4) (m - 5) = - 8

m = 4, m = - 3

Solution steps

(m + 4) (m - 5) = - 8

Expand

(m + 4) (m - 5) : m^{2} - m - 20

m^{2} - m - 20 = - 8

Move

8

to the left side

m^{2} - m - 12 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 12) = 7

m_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 1}, m_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 1}

m = \frac{- ( - 1 ) + 7}{2 \cdot 1} : 4

m = \frac{- ( - 1 ) - 7}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

m = 4, m = - 3

6 t^{2} - 12 t = 0

v^{2} - 2 v - 24 = 0

y^{2} = - 10 y

r^{2} = (- 2)^{2} + (23)^{2}

\frac{x ^{2}}{3} + 2 x = x

What are the solutions to the equation (m+4)(m-5)=-8 ?
The solutions to the equation (m+4)(m-5)=-8 are m=4,m=-3
Find the zeros of (m+4)(m-5)=-8
The zeros of (m+4)(m-5)=-8 are m=4,m=-3