What are the solutions to the equation 2m^2-7m-15=0,(m-5)(2m+3)=0 ?

The solutions to the equation 2m^2-7m-15=0,(m-5)(2m+3)=0 are m=5,m=-3/2

Find the zeros of 2m^2-7m-15=0,(m-5)(2m+3)=0

The zeros of 2m^2-7m-15=0,(m-5)(2m+3)=0 are m=5,m=-3/2

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2 m^{2} - 7 m - 15 = 0, (m - 5) (2 m + 3) = 0

m = 5, m = - \frac{3}{2}

Decimal Notation

m = 5, m = - 1.5

Solution steps

2 m^{2} - 7 m - 15 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 15) = 13

m_{1, 2} = \frac{- ( - 7 ) \pm 13}{2 \cdot 2}

Separate the solutions

m_{1} = \frac{- ( - 7 ) + 13}{2 \cdot 2}, m_{2} = \frac{- ( - 7 ) - 13}{2 \cdot 2}

m = \frac{- ( - 7 ) + 13}{2 \cdot 2} : 5

m = \frac{- ( - 7 ) - 13}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

m = 5, m = - \frac{3}{2}

2 p + p^{2} = 1358

x^{2} + (m - 1) \cdot x + 2 m - 6 = 0

x^{2} = - 3, x^{1} = 7

25 m^{2} - 54 m + 128 = 0

a, a^{2} + b^{2} - 2 (3 a + 2 b) + 13 = 0

What are the solutions to the equation 2m^2-7m-15=0,(m-5)(2m+3)=0 ?
The solutions to the equation 2m^2-7m-15=0,(m-5)(2m+3)=0 are m=5,m=-3/2
Find the zeros of 2m^2-7m-15=0,(m-5)(2m+3)=0
The zeros of 2m^2-7m-15=0,(m-5)(2m+3)=0 are m=5,m=-3/2