What are the solutions to the equation (0.8-x)(0.7-x)-0.06=0 ?

The solutions to the equation (0.8-x)(0.7-x)-0.06=0 are x=(1.5+sqrt(0.25))/2 ,x=(1.5-sqrt(0.25))/2

Find the zeros of (0.8-x)(0.7-x)-0.06=0

The zeros of (0.8-x)(0.7-x)-0.06=0 are x=(1.5+sqrt(0.25))/2 ,x=(1.5-sqrt(0.25))/2

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(0.8 - x) (0.7 - x) - 0.06 = 0

x = \frac{1.5 + 0.25}{2}, x = \frac{1.5 - 0.25}{2}

Decimal Notation

x = 1, x = 0.5

Solution steps

(0.8 - x) (0.7 - x) - 0.06 = 0

Expand

(0.8 - x) (0.7 - x) - 0.06 : x^{2} - 1.5 x + 0.5

x^{2} - 1.5 x + 0.5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1.5 ) \pm ( - 1.5 ) ^{2} - 4 \cdot 1 \cdot 0.5}{2 \cdot 1}

(- 1.5)^{2} - 4 \cdot 1 \cdot 0.5 = 0.25

x_{1, 2} = \frac{- ( - 1.5 ) \pm 0.25}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1.5 ) + 0.25}{2 \cdot 1}, x_{2} = \frac{- ( - 1.5 ) - 0.25}{2 \cdot 1}

x = \frac{- ( - 1.5 ) + 0.25}{2 \cdot 1} : \frac{1.5 + 0.25}{2}

x = \frac{- ( - 1.5 ) - 0.25}{2 \cdot 1} : \frac{1.5 - 0.25}{2}

The solutions to the quadratic equation are:

x = \frac{1.5 + 0.25}{2}, x = \frac{1.5 - 0.25}{2}

c^{2} + 11 c + 18

\frac{3 + t ^{4}}{u t ^{2} + u ^{4} t ^{2}}

18 \geq - 2 x

2^{- \frac{8}{5}} \cdot 1 6^{- \frac{8}{5}}

8 x^{2} - 4

What are the solutions to the equation (0.8-x)(0.7-x)-0.06=0 ?
The solutions to the equation (0.8-x)(0.7-x)-0.06=0 are x=(1.5+sqrt(0.25))/2 ,x=(1.5-sqrt(0.25))/2
Find the zeros of (0.8-x)(0.7-x)-0.06=0
The zeros of (0.8-x)(0.7-x)-0.06=0 are x=(1.5+sqrt(0.25))/2 ,x=(1.5-sqrt(0.25))/2