What are the solutions to the equation 2x^2+3x-5=4 ?

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

Find the zeros of 2x^2+3x-5=4

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

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2 x^{2} + 3 x - 5 = 4

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

Decimal Notation

x = 1.5, x = - 3

Solution steps

2 x^{2} + 3 x - 5 = 4

Move

4

to the left side

2 x^{2} + 3 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 9) = 9

x_{1, 2} = \frac{- 3 \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 9}{2 \cdot 2}, x_{2} = \frac{- 3 - 9}{2 \cdot 2}

x = \frac{- 3 + 9}{2 \cdot 2} : \frac{3}{2}

x = \frac{- 3 - 9}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

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

5 x - 7 \geq - 17

\frac{1}{2} + \frac{1}{7}

x^{2} + 6 x < 0

\frac{20 !}{5 ! ( 20 - 5 ) !}

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