Solve (2)(t+sqrt{t})(t-sqrt{t})+(t-2)^2=7-5t

$$( 2 ) ( t + \sqrt { t } ) ( t - \sqrt { t } ) + ( t - 2 ) ^ { 2 } = 7 - 5 t$$

Solve for t (complex solution)

$$t = \frac{\sqrt{37} + 1}{6} \approx 1.180460422$$

$$t=\frac{1-\sqrt{37}}{6}\approx -0.847127088$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by t+\sqrt{t}.

$$\left(2t+2\sqrt{t}\right)\left(t-\sqrt{t}\right)+\left(t-2\right)^{2}=7-5t$$

Use the distributive property to multiply 2t+2\sqrt{t} by t-\sqrt{t} and combine like terms.

$$2t^{2}-2\left(\sqrt{t}\right)^{2}+\left(t-2\right)^{2}=7-5t$$

Calculate \sqrt{t} to the power of 2 and get t.

$$2t^{2}-2t+\left(t-2\right)^{2}=7-5t$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2\right)^{2}.

$$2t^{2}-2t+t^{2}-4t+4=7-5t$$

Combine 2t^{2} and t^{2} to get 3t^{2}.

$$3t^{2}-2t-4t+4=7-5t$$

Combine -2t and -4t to get -6t.

$$3t^{2}-6t+4=7-5t$$

Subtract 7 from both sides.

$$3t^{2}-6t+4-7=-5t$$

Subtract 7 from 4 to get -3.

$$3t^{2}-6t-3=-5t$$

Add 5t to both sides.

$$3t^{2}-6t-3+5t=0$$

Combine -6t and 5t to get -t.

$$3t^{2}-t-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-1\right)±\sqrt{1-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$t=\frac{-\left(-1\right)±\sqrt{1+36}}{2\times 3}$$

Add 1 to 36.

$$t=\frac{-\left(-1\right)±\sqrt{37}}{2\times 3}$$

The opposite of -1 is 1.

$$t=\frac{1±\sqrt{37}}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{1±\sqrt{37}}{6}$$

Now solve the equation t=\frac{1±\sqrt{37}}{6} when ± is plus. Add 1 to \sqrt{37}.

$$t=\frac{\sqrt{37}+1}{6}$$

Now solve the equation t=\frac{1±\sqrt{37}}{6} when ± is minus. Subtract \sqrt{37} from 1.

$$t=\frac{1-\sqrt{37}}{6}$$

The equation is now solved.

$$t=\frac{\sqrt{37}+1}{6} t=\frac{1-\sqrt{37}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by t+\sqrt{t}.

$$\left(2t+2\sqrt{t}\right)\left(t-\sqrt{t}\right)+\left(t-2\right)^{2}=7-5t$$

Use the distributive property to multiply 2t+2\sqrt{t} by t-\sqrt{t} and combine like terms.

$$2t^{2}-2\left(\sqrt{t}\right)^{2}+\left(t-2\right)^{2}=7-5t$$

Calculate \sqrt{t} to the power of 2 and get t.

$$2t^{2}-2t+\left(t-2\right)^{2}=7-5t$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2\right)^{2}.

$$2t^{2}-2t+t^{2}-4t+4=7-5t$$

Combine 2t^{2} and t^{2} to get 3t^{2}.

$$3t^{2}-2t-4t+4=7-5t$$

Combine -2t and -4t to get -6t.

$$3t^{2}-6t+4=7-5t$$

Add 5t to both sides.

$$3t^{2}-6t+4+5t=7$$

Combine -6t and 5t to get -t.

$$3t^{2}-t+4=7$$

Subtract 4 from both sides.

$$3t^{2}-t=7-4$$

Subtract 4 from 7 to get 3.

$$3t^{2}-t=3$$

Divide both sides by 3.

$$\frac{3t^{2}-t}{3}=\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}-\frac{1}{3}t=\frac{3}{3}$$

Divide 3 by 3.

$$t^{2}-\frac{1}{3}t=1$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{1}{3}t+\left(-\frac{1}{6}\right)^{2}=1+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1}{3}t+\frac{1}{36}=1+\frac{1}{36}$$

Add 1 to \frac{1}{36}.

$$t^{2}-\frac{1}{3}t+\frac{1}{36}=\frac{37}{36}$$

Factor t^{2}-\frac{1}{3}t+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{6}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{6}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$t-\frac{1}{6}=\frac{\sqrt{37}}{6} t-\frac{1}{6}=-\frac{\sqrt{37}}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$t=\frac{\sqrt{37}+1}{6} t=\frac{1-\sqrt{37}}{6}$$

Solve for t

$$t = \frac{\sqrt{37} + 1}{6} \approx 1.180460422$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by t+\sqrt{t}.

$$\left(2t+2\sqrt{t}\right)\left(t-\sqrt{t}\right)+\left(t-2\right)^{2}=7-5t$$

Use the distributive property to multiply 2t+2\sqrt{t} by t-\sqrt{t} and combine like terms.

$$2t^{2}-2\left(\sqrt{t}\right)^{2}+\left(t-2\right)^{2}=7-5t$$

Calculate \sqrt{t} to the power of 2 and get t.

$$2t^{2}-2t+\left(t-2\right)^{2}=7-5t$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2\right)^{2}.

$$2t^{2}-2t+t^{2}-4t+4=7-5t$$

Combine 2t^{2} and t^{2} to get 3t^{2}.

$$3t^{2}-2t-4t+4=7-5t$$

Combine -2t and -4t to get -6t.

$$3t^{2}-6t+4=7-5t$$

Subtract 7 from both sides.

$$3t^{2}-6t+4-7=-5t$$

Subtract 7 from 4 to get -3.

$$3t^{2}-6t-3=-5t$$

Add 5t to both sides.

$$3t^{2}-6t-3+5t=0$$

Combine -6t and 5t to get -t.

$$3t^{2}-t-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-1\right)±\sqrt{1-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$t=\frac{-\left(-1\right)±\sqrt{1+36}}{2\times 3}$$

Add 1 to 36.

$$t=\frac{-\left(-1\right)±\sqrt{37}}{2\times 3}$$

The opposite of -1 is 1.

$$t=\frac{1±\sqrt{37}}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{1±\sqrt{37}}{6}$$

Now solve the equation t=\frac{1±\sqrt{37}}{6} when ± is plus. Add 1 to \sqrt{37}.

$$t=\frac{\sqrt{37}+1}{6}$$

Now solve the equation t=\frac{1±\sqrt{37}}{6} when ± is minus. Subtract \sqrt{37} from 1.

$$t=\frac{1-\sqrt{37}}{6}$$

The equation is now solved.

$$t=\frac{\sqrt{37}+1}{6} t=\frac{1-\sqrt{37}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by t+\sqrt{t}.

$$\left(2t+2\sqrt{t}\right)\left(t-\sqrt{t}\right)+\left(t-2\right)^{2}=7-5t$$

Use the distributive property to multiply 2t+2\sqrt{t} by t-\sqrt{t} and combine like terms.

$$2t^{2}-2\left(\sqrt{t}\right)^{2}+\left(t-2\right)^{2}=7-5t$$

Calculate \sqrt{t} to the power of 2 and get t.

$$2t^{2}-2t+\left(t-2\right)^{2}=7-5t$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2\right)^{2}.

$$2t^{2}-2t+t^{2}-4t+4=7-5t$$

Combine 2t^{2} and t^{2} to get 3t^{2}.

$$3t^{2}-2t-4t+4=7-5t$$

Combine -2t and -4t to get -6t.

$$3t^{2}-6t+4=7-5t$$

Add 5t to both sides.

$$3t^{2}-6t+4+5t=7$$

Combine -6t and 5t to get -t.

$$3t^{2}-t+4=7$$

Subtract 4 from both sides.

$$3t^{2}-t=7-4$$

Subtract 4 from 7 to get 3.

$$3t^{2}-t=3$$

Divide both sides by 3.

$$\frac{3t^{2}-t}{3}=\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}-\frac{1}{3}t=\frac{3}{3}$$

Divide 3 by 3.

$$t^{2}-\frac{1}{3}t=1$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{1}{3}t+\left(-\frac{1}{6}\right)^{2}=1+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1}{3}t+\frac{1}{36}=1+\frac{1}{36}$$

Add 1 to \frac{1}{36}.

$$t^{2}-\frac{1}{3}t+\frac{1}{36}=\frac{37}{36}$$

Factor t^{2}-\frac{1}{3}t+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{6}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{6}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$t-\frac{1}{6}=\frac{\sqrt{37}}{6} t-\frac{1}{6}=-\frac{\sqrt{37}}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$t=\frac{\sqrt{37}+1}{6} t=\frac{1-\sqrt{37}}{6}$$