What are the solutions to the equation (m+4)(m-1)=3 ?

The solutions to the equation (m+4)(m-1)=3 are m=(-3+sqrt(37))/2 ,m=(-3-sqrt(37))/2

Find the zeros of (m+4)(m-1)=3

The zeros of (m+4)(m-1)=3 are m=(-3+sqrt(37))/2 ,m=(-3-sqrt(37))/2

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(m + 4) (m - 1) = 3

m = \frac{- 3 + 37}{2}, m = \frac{- 3 - 37}{2}

Decimal Notation

m = 1.54138 \dots, m = - 4.54138 \dots

Solution steps

(m + 4) (m - 1) = 3

Expand

(m + 4) (m - 1) : m^{2} + 3 m - 4

m^{2} + 3 m - 4 = 3

Move

3

to the left side

m^{2} + 3 m - 7 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 7) = 37

m_{1, 2} = \frac{- 3 \pm 37}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 3 + 37}{2 \cdot 1}, m_{2} = \frac{- 3 - 37}{2 \cdot 1}

m = \frac{- 3 + 37}{2 \cdot 1} : \frac{- 3 + 37}{2}

m = \frac{- 3 - 37}{2 \cdot 1} : \frac{- 3 - 37}{2}

The solutions to the quadratic equation are:

m = \frac{- 3 + 37}{2}, m = \frac{- 3 - 37}{2}

(2 x - 1)^{2} = 4 x

- 5 x^{2} = 90 x + 360

3 r^{2} - 2 r = 21

1 2^{2} + x^{2} = 2 4^{2}

y (4 y + 1) = 8

What are the solutions to the equation (m+4)(m-1)=3 ?
The solutions to the equation (m+4)(m-1)=3 are m=(-3+sqrt(37))/2 ,m=(-3-sqrt(37))/2
Find the zeros of (m+4)(m-1)=3
The zeros of (m+4)(m-1)=3 are m=(-3+sqrt(37))/2 ,m=(-3-sqrt(37))/2