What are the solutions to the equation 2x^2+3x-16=4 ?

The solutions to the equation 2x^2+3x-16=4 are x= 5/2 ,x=-4

Find the zeros of 2x^2+3x-16=4

The zeros of 2x^2+3x-16=4 are x= 5/2 ,x=-4

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2 x^{2} + 3 x - 16 = 4

x = \frac{5}{2}, x = - 4

Decimal Notation

x = 2.5, x = - 4

Solution steps

2 x^{2} + 3 x - 16 = 4

Move

4

to the left side

2 x^{2} + 3 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 20) = 13

x_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 13}{2 \cdot 2}, x_{2} = \frac{- 3 - 13}{2 \cdot 2}

x = \frac{- 3 + 13}{2 \cdot 2} : \frac{5}{2}

x = \frac{- 3 - 13}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - 4

v^{2} = 47

6 x^{2} = - x + 1

(x - 1)^{2} + 3 = 2 x + 1

64 = 80 t - 16 t^{2}

2 x^{2} + x = - 9

What are the solutions to the equation 2x^2+3x-16=4 ?
The solutions to the equation 2x^2+3x-16=4 are x= 5/2 ,x=-4
Find the zeros of 2x^2+3x-16=4
The zeros of 2x^2+3x-16=4 are x= 5/2 ,x=-4