What are the solutions to the equation 2x^2-3x+1=38 ?

The solutions to the equation 2x^2-3x+1=38 are x=(3+sqrt(305))/4 ,x=(3-sqrt(305))/4

Find the zeros of 2x^2-3x+1=38

The zeros of 2x^2-3x+1=38 are x=(3+sqrt(305))/4 ,x=(3-sqrt(305))/4

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2 x^{2} - 3 x + 1 = 38

x = \frac{3 + 305}{4}, x = \frac{3 - 305}{4}

Decimal Notation

x = 5.11606 \dots, x = - 3.61606 \dots

Solution steps

2 x^{2} - 3 x + 1 = 38

Move

38

to the left side

2 x^{2} - 3 x - 37 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 37 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 37) = 305

x_{1, 2} = \frac{- ( - 3 ) \pm 305}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 305}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 305}{2 \cdot 2}

x = \frac{- ( - 3 ) + 305}{2 \cdot 2} : \frac{3 + 305}{4}

x = \frac{- ( - 3 ) - 305}{2 \cdot 2} : \frac{3 - 305}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 305}{4}, x = \frac{3 - 305}{4}

y, x^{2} + y^{2} = C e^{2 x}

t^{2} - 5 = 3

0.15 x^{2} = 0.6

x^{2} + 8 x = 0

- y^{2} + 6 y - 12 = 0

What are the solutions to the equation 2x^2-3x+1=38 ?
The solutions to the equation 2x^2-3x+1=38 are x=(3+sqrt(305))/4 ,x=(3-sqrt(305))/4
Find the zeros of 2x^2-3x+1=38
The zeros of 2x^2-3x+1=38 are x=(3+sqrt(305))/4 ,x=(3-sqrt(305))/4