(-2t^2-7t+5)+(-8t^2+4t-3)-এর সমাধান করুন

$$( - 2 t ^ { 2 } - 7 t + 5 ) + ( - 8 t ^ { 2 } + 4 t - 3 )$$

Evaluate

$$2-3t-10t^{2}$$

Solution Steps

Combine -2t^{2} and -8t^{2} to get -10t^{2}.

$$-10t^{2}-7t+5+4t-3$$

Combine -7t and 4t to get -3t.

$$-10t^{2}-3t+5-3$$

Subtract 3 from 5 to get 2.

$$-10t^{2}-3t+2$$

Factor

$$-10\left(t-\frac{-\sqrt{89}-3}{20}\right)\left(t-\frac{\sqrt{89}-3}{20}\right)$$

Steps Using the Quadratic Formula

Combine -2t^{2} and -8t^{2} to get -10t^{2}.

$$factor(-10t^{2}-7t+5+4t-3)$$

Combine -7t and 4t to get -3t.

$$factor(-10t^{2}-3t+5-3)$$

Subtract 3 from 5 to get 2.

$$factor(-10t^{2}-3t+2)$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-10t^{2}-3t+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)\times 2}}{2\left(-10\right)}$$

Square -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)\times 2}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$t=\frac{-\left(-3\right)±\sqrt{9+40\times 2}}{2\left(-10\right)}$$

Multiply 40 times 2.

$$t=\frac{-\left(-3\right)±\sqrt{9+80}}{2\left(-10\right)}$$

Add 9 to 80.

$$t=\frac{-\left(-3\right)±\sqrt{89}}{2\left(-10\right)}$$

The opposite of -3 is 3.

$$t=\frac{3±\sqrt{89}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$t=\frac{3±\sqrt{89}}{-20}$$

Now solve the equation t=\frac{3±\sqrt{89}}{-20} when ± is plus. Add 3 to \sqrt{89}.

$$t=\frac{\sqrt{89}+3}{-20}$$

Divide 3+\sqrt{89} by -20.

$$t=\frac{-\sqrt{89}-3}{20}$$

Now solve the equation t=\frac{3±\sqrt{89}}{-20} when ± is minus. Subtract \sqrt{89} from 3.

$$t=\frac{3-\sqrt{89}}{-20}$$

Divide 3-\sqrt{89} by -20.

$$t=\frac{\sqrt{89}-3}{20}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3-\sqrt{89}}{20} for x_{1} and \frac{-3+\sqrt{89}}{20} for x_{2}.

$$-10t^{2}-3t+2=-10\left(t-\frac{-\sqrt{89}-3}{20}\right)\left(t-\frac{\sqrt{89}-3}{20}\right)$$