What are the solutions to the equation 2m^2+7m+43=52 ?

The solutions to the equation 2m^2+7m+43=52 are m=1,m=-9/2

Find the zeros of 2m^2+7m+43=52

The zeros of 2m^2+7m+43=52 are m=1,m=-9/2

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2 m^{2} + 7 m + 43 = 52

m = 1, m = - \frac{9}{2}

Decimal Notation

m = 1, m = - 4.5

Solution steps

2 m^{2} + 7 m + 43 = 52

Move

52

to the left side

2 m^{2} + 7 m - 9 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 9) = 11

m_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 2}

Separate the solutions

m_{1} = \frac{- 7 + 11}{2 \cdot 2}, m_{2} = \frac{- 7 - 11}{2 \cdot 2}

m = \frac{- 7 + 11}{2 \cdot 2} : 1

m = \frac{- 7 - 11}{2 \cdot 2} : - \frac{9}{2}

The solutions to the quadratic equation are:

m = 1, m = - \frac{9}{2}

x^{2} - \frac{25}{28} x + \frac{3}{28} = 0

y^{2} - 7 y = - 10

5 y^{2} = - 8 y - 3

4 x^{2} + 4 x^{2} = 7

- 4 x^{2} + 13 x - 3 = 0

What are the solutions to the equation 2m^2+7m+43=52 ?
The solutions to the equation 2m^2+7m+43=52 are m=1,m=-9/2
Find the zeros of 2m^2+7m+43=52
The zeros of 2m^2+7m+43=52 are m=1,m=-9/2