What are the solutions to the equation 0=0.4x^2-43x+450 ?

The solutions to the equation 0=0.4x^2-43x+450 are x=(5(43+sqrt(1129)))/4 ,x=(5(43-sqrt(1129)))/4

Find the zeros of 0=0.4x^2-43x+450

The zeros of 0=0.4x^2-43x+450 are x=(5(43+sqrt(1129)))/4 ,x=(5(43-sqrt(1129)))/4

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0 = 0.4 x^{2} - 43 x + 450

x = \frac{5 ( 43 + 1129 )}{4}, x = \frac{5 ( 43 - 1129 )}{4}

Decimal Notation

x = 95.75074 \dots, x = 11.74925 \dots

Solution steps

0 = 0.4 x^{2} - 43 x + 450

Multiply both sides by

10

0 = 4 x^{2} - 430 x + 4500

Switch sides

4 x^{2} - 430 x + 4500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 430 ) \pm ( - 430 ) ^{2} - 4 \cdot 4 \cdot 4500}{2 \cdot 4}

(- 430)^{2} - 4 \cdot 4 \cdot 4500 = 101129

x_{1, 2} = \frac{- ( - 430 ) \pm 10 1129}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 430 ) + 10 1129}{2 \cdot 4}, x_{2} = \frac{- ( - 430 ) - 10 1129}{2 \cdot 4}

x = \frac{- ( - 430 ) + 10 1129}{2 \cdot 4} : \frac{5 ( 43 + 1129 )}{4}

x = \frac{- ( - 430 ) - 10 1129}{2 \cdot 4} : \frac{5 ( 43 - 1129 )}{4}

The solutions to the quadratic equation are:

x = \frac{5 ( 43 + 1129 )}{4}, x = \frac{5 ( 43 - 1129 )}{4}

7 x + 2 x + 27 = 180

\frac{7 ^{\frac{2}{3}}}{7 ^{\frac{5}{3}}}

10 - 8

x + 2 \geq - 1

\frac{t}{2} - 1 \geq 1

What are the solutions to the equation 0=0.4x^2-43x+450 ?
The solutions to the equation 0=0.4x^2-43x+450 are x=(5(43+sqrt(1129)))/4 ,x=(5(43-sqrt(1129)))/4
Find the zeros of 0=0.4x^2-43x+450
The zeros of 0=0.4x^2-43x+450 are x=(5(43+sqrt(1129)))/4 ,x=(5(43-sqrt(1129)))/4