What are the solutions to the equation m^2=15m-56 ?

The solutions to the equation m^2=15m-56 are m=8,m=7

Find the zeros of m^2=15m-56

The zeros of m^2=15m-56 are m=8,m=7

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m^{2} = 15 m - 56

m = 8, m = 7

Solution steps

m^{2} = 15 m - 56

Move

56

to the left side

m^{2} + 56 = 15 m

Move

15 m

to the left side

m^{2} + 56 - 15 m = 0

Write in the standard form

a x^{2} + b x + c = 0

m^{2} - 15 m + 56 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 56 = 1

m_{1, 2} = \frac{- ( - 15 ) \pm 1}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 15 ) + 1}{2 \cdot 1}, m_{2} = \frac{- ( - 15 ) - 1}{2 \cdot 1}

m = \frac{- ( - 15 ) + 1}{2 \cdot 1} : 8

m = \frac{- ( - 15 ) - 1}{2 \cdot 1} : 7

The solutions to the quadratic equation are:

m = 8, m = 7

115 = - 15 t^{2} + 120 t + 10

a^{2} - 4 a

(- \frac{1}{3})^{3}

2 (x + 7)^{2} = 36

e^{- i (\frac{π}{2})}

What are the solutions to the equation m^2=15m-56 ?
The solutions to the equation m^2=15m-56 are m=8,m=7
Find the zeros of m^2=15m-56
The zeros of m^2=15m-56 are m=8,m=7