What are the solutions to the equation 2x^2-3x=20 ?

The solutions to the equation 2x^2-3x=20 are x=4,x=-5/2

Find the zeros of 2x^2-3x=20

The zeros of 2x^2-3x=20 are x=4,x=-5/2

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2 x^{2} - 3 x = 20

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

Solution steps

2 x^{2} - 3 x = 20

Move

20

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} : 4

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

The solutions to the quadratic equation are:

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

\frac{( 2 )}{x} \leq x

lo g_{2} (16)

5 + \frac{1}{2}

\frac{3}{48}

2 a x + 2 b x + a y + b y

What are the solutions to the equation 2x^2-3x=20 ?
The solutions to the equation 2x^2-3x=20 are x=4,x=-5/2
Find the zeros of 2x^2-3x=20
The zeros of 2x^2-3x=20 are x=4,x=-5/2