What is the number and type of roots for the equation 7x^2+8x-12=2x^2+14x-4?

I hope this helps you

7x^2+8x-12-2x^2-14x+4=0

5x^2-6x-8=0

a=5 b= -6 c= -8

disctirminant =b^2-4ac

disctirminant =(-6)^2-4.5. (-8)

disctirminant =36+160

disctirminant =196

x1=6+ square root of 196/2.5

x1=3+2 square root of 3/5

x2=3- square root of 3/5

7x^2+8x-12=2x^2+14x-4

⇒ 7x^2 - 2x^2 + 8x - 14x -12 +4 = 0

⇒ 5x^2 - 6x - 8 = 0

b^2 - 4ac = (-6)^2 - 4 * 5 * (-8) = 36 + 160 = 196 > 0

So there are two real roots.

Hope it helps!

Square Root of 41 to the nearest tenth, means to calculate the square root of 41 where the answer should only have one number after the decimal point.

Here are step-by-step instructions for how to get the square root of 41 to the nearest tenth:

Step 1: Calculate

We calculate the square root of 41 to be:

√41 = 6.40312423743285

Step 2: Reduce

Reduce the tail of the answer above to two numbers after the decimal point:

6.40

Step 3: Round

Round 6.40 so you only have one digit after the decimal point to get the

6.4

To check that the answer is correct, use your calculator to confirm that 6.42 is about 41.

Step-by-step explanation: