What are the solutions to the equation 8x^2-20x+12=4 ?

The solutions to the equation 8x^2-20x+12=4 are x=2,x= 1/2

Find the zeros of 8x^2-20x+12=4

The zeros of 8x^2-20x+12=4 are x=2,x= 1/2

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8 x^{2} - 20 x + 12 = 4

x = 2, x = \frac{1}{2}

Decimal Notation

x = 2, x = 0.5

Solution steps

8 x^{2} - 20 x + 12 = 4

Move

4

to the left side

8 x^{2} - 20 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 8 \cdot 8}{2 \cdot 8}

(- 20)^{2} - 4 \cdot 8 \cdot 8 = 12

x_{1, 2} = \frac{- ( - 20 ) \pm 12}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 12}{2 \cdot 8}, x_{2} = \frac{- ( - 20 ) - 12}{2 \cdot 8}

x = \frac{- ( - 20 ) + 12}{2 \cdot 8} : 2

x = \frac{- ( - 20 ) - 12}{2 \cdot 8} : \frac{1}{2}

The solutions to the quadratic equation are:

x = 2, x = \frac{1}{2}

x, x^{2} + y^{3} = 2 x^{2} y

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

62 x^{2} + 60 x - 30 = 6770

y, x^{2} + (y - x)^{2} = 1

- x^{2} + 1.5 x + 5 = - 5

What are the solutions to the equation 8x^2-20x+12=4 ?
The solutions to the equation 8x^2-20x+12=4 are x=2,x= 1/2
Find the zeros of 8x^2-20x+12=4
The zeros of 8x^2-20x+12=4 are x=2,x= 1/2