What are the solutions to the equation 3x^2-40x+150=73 ?

The solutions to the equation 3x^2-40x+150=73 are x=11,x= 7/3

Find the zeros of 3x^2-40x+150=73

The zeros of 3x^2-40x+150=73 are x=11,x= 7/3

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3 x^{2} - 40 x + 150 = 73

x = 11, x = \frac{7}{3}

Decimal Notation

x = 11, x = 2.33333 \dots

Solution steps

3 x^{2} - 40 x + 150 = 73

Move

73

to the left side

3 x^{2} - 40 x + 77 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 3 \cdot 77}{2 \cdot 3}

(- 40)^{2} - 4 \cdot 3 \cdot 77 = 26

x_{1, 2} = \frac{- ( - 40 ) \pm 26}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 26}{2 \cdot 3}, x_{2} = \frac{- ( - 40 ) - 26}{2 \cdot 3}

x = \frac{- ( - 40 ) + 26}{2 \cdot 3} : 11

x = \frac{- ( - 40 ) - 26}{2 \cdot 3} : \frac{7}{3}

The solutions to the quadratic equation are:

x = 11, x = \frac{7}{3}

x^{2} = 3 x - 3

14 x^{2} - 31 x + 15 = 0

(x - 4)^{2} = 12

21000 = 0.68 \cdot 22 \cdot x \cdot \frac{2.5}{1.4} \cdot (54 - 0.4 x)

\frac{y ^{2}}{2} - 3 = y + 1

What are the solutions to the equation 3x^2-40x+150=73 ?
The solutions to the equation 3x^2-40x+150=73 are x=11,x= 7/3
Find the zeros of 3x^2-40x+150=73
The zeros of 3x^2-40x+150=73 are x=11,x= 7/3