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

The solutions to the equation 4x^2-3x=2 are x=(3+sqrt(41))/8 ,x=(3-sqrt(41))/8

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

The zeros of 4x^2-3x=2 are x=(3+sqrt(41))/8 ,x=(3-sqrt(41))/8

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4 x^{2} - 3 x = 2

x = \frac{3 + 41}{8}, x = \frac{3 - 41}{8}

Decimal Notation

x = 1.17539 \dots, x = - 0.42539 \dots

Solution steps

4 x^{2} - 3 x = 2

Move

2

to the left side

4 x^{2} - 3 x - 2 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 4 (- 2) = 41

x_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 4}, x_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 4}

x = \frac{- ( - 3 ) + 41}{2 \cdot 4} : \frac{3 + 41}{8}

x = \frac{- ( - 3 ) - 41}{2 \cdot 4} : \frac{3 - 41}{8}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{8}, x = \frac{3 - 41}{8}

(10 - x) (- 12 - x) + 40 = 0

x (4 x - 1) = 3 x (x - 2) + 6

x, x^{2} - k x + 16 = 0

x^{2} = - 1 - 4 x

3 x^{2} - 15 = 4 x

What are the solutions to the equation 4x^2-3x=2 ?
The solutions to the equation 4x^2-3x=2 are x=(3+sqrt(41))/8 ,x=(3-sqrt(41))/8
Find the zeros of 4x^2-3x=2
The zeros of 4x^2-3x=2 are x=(3+sqrt(41))/8 ,x=(3-sqrt(41))/8