What are the solutions to the equation x^2-3x+2=6*0.7 ?

The solutions to the equation x^2-3x+2=6*0.7 are x=(15+sqrt(445))/(10),x=(15-sqrt(445))/(10)

Find the zeros of x^2-3x+2=6*0.7

The zeros of x^2-3x+2=6*0.7 are x=(15+sqrt(445))/(10),x=(15-sqrt(445))/(10)

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x^{2} - 3 x + 2 = 6 \cdot 0.7

x = \frac{15 + 445}{10}, x = \frac{15 - 445}{10}

Decimal Notation

x = 3.60950 \dots, x = - 0.60950 \dots

Solution steps

x^{2} - 3 x + 2 = 6 \cdot 0.7

Multiply both sides by

10

10 x^{2} - 30 x + 20 = 42

Move

42

to the left side

10 x^{2} - 30 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 10 ( - 22 )}{2 \cdot 10}

(- 30)^{2} - 4 \cdot 10 (- 22) = 2445

x_{1, 2} = \frac{- ( - 30 ) \pm 2 445}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 30 ) + 2 445}{2 \cdot 10}, x_{2} = \frac{- ( - 30 ) - 2 445}{2 \cdot 10}

x = \frac{- ( - 30 ) + 2 445}{2 \cdot 10} : \frac{15 + 445}{10}

x = \frac{- ( - 30 ) - 2 445}{2 \cdot 10} : \frac{15 - 445}{10}

The solutions to the quadratic equation are:

x = \frac{15 + 445}{10}, x = \frac{15 - 445}{10}

x - 3 x^{2} < ∣ 1 - 3 x ∣

246 = 77 + 13 x

u^{2} - 9 u - 2 u + 18

60 = \frac{3}{4} h

(- 3 v)^{5}

What are the solutions to the equation x^2-3x+2=6*0.7 ?
The solutions to the equation x^2-3x+2=6*0.7 are x=(15+sqrt(445))/(10),x=(15-sqrt(445))/(10)
Find the zeros of x^2-3x+2=6*0.7
The zeros of x^2-3x+2=6*0.7 are x=(15+sqrt(445))/(10),x=(15-sqrt(445))/(10)