What are the solutions to the equation 15=2x^2-7x ?

The solutions to the equation 15=2x^2-7x are x=5,x=-3/2

Find the zeros of 15=2x^2-7x

The zeros of 15=2x^2-7x are x=5,x=-3/2

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15 = 2 x^{2} - 7 x

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

Decimal Notation

x = 5, x = - 1.5

Solution steps

15 = 2 x^{2} - 7 x

Switch sides

2 x^{2} - 7 x = 15

Move

15

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(1 + \frac{0.11}{4})^{4} - 1

3 (x + 4) \geq 2 (x - 5) + 1

1990 (6) = 4 x^{2} + 5 x + 2

2 x^{2} - 24 x + 80 = 0

48 n^{3}

What are the solutions to the equation 15=2x^2-7x ?
The solutions to the equation 15=2x^2-7x are x=5,x=-3/2
Find the zeros of 15=2x^2-7x
The zeros of 15=2x^2-7x are x=5,x=-3/2