Whakaoti i te left(6-aright)x-20=-0.05x^2+10x-40

$$\left( 6-a \right) x-20 = -0.05 { x }^{ 2 } +10x-40$$

Whakaoti mō a

$$a=\frac{x}{20}-4+\frac{20}{x}$$

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 6-a ki te x.

$$6x-ax-20=-0.05x^{2}+10x-40$$

Tangohia te 6x mai i ngā taha e rua.

$$-ax-20=-0.05x^{2}+10x-40-6x$$

Pahekotia te 10x me -6x, ka 4x.

$$-ax-20=-0.05x^{2}+4x-40$$

Me tāpiri te 20 ki ngā taha e rua.

$$-ax=-0.05x^{2}+4x-40+20$$

Tāpirihia te -40 ki te 20, ka -20.

$$-ax=-0.05x^{2}+4x-20$$

He hanga arowhānui tō te whārite.

$$\left(-x\right)a=-\frac{x^{2}}{20}+4x-20$$

Whakawehea ngā taha e rua ki te -x.

$$\frac{\left(-x\right)a}{-x}=\frac{-\frac{x^{2}}{20}+4x-20}{-x}$$

Mā te whakawehe ki te -x ka wetekia te whakareanga ki te -x.

$$a=\frac{-\frac{x^{2}}{20}+4x-20}{-x}$$

Whakawehe -\frac{x^{2}}{20}+4x-20 ki te -x.

$$a=\frac{x}{20}-4+\frac{20}{x}$$

Whakaoti mō x (complex solution)

$$x=10\sqrt{\left(a+2\right)\left(a+6\right)}+10a+40$$

$$x=-10\sqrt{\left(a+2\right)\left(a+6\right)}+10a+40$$

Whakaoti mō x

$$x=10\left(\sqrt{\left(a+2\right)\left(a+6\right)}+a+4\right)$$

$$x=10\left(-\sqrt{\left(a+2\right)\left(a+6\right)}+a+4\right)\text{, }a\leq -6\text{ or }a\geq -2$$