Question

𝔸𝕝𝕘𝕖𝕓𝕣𝕒 𝕀𝕀: ℚ𝕦𝕒𝕕𝕣𝕒𝕥𝕚𝕔 𝔽𝕠𝕣𝕞𝕦𝕝𝕒
🅃🅄🅃🄾🅁🄸🄰🄻:

Answer 1

𝘘𝘶𝘦𝘴𝘵𝘪𝘰𝘯: Solve this equation by Factoring.
\[2x^2-5=3x\]
Now, we want to move the 3x to the other side. Subtract the 3x from both sides.
\[2x^2-3x-5=0\]
We always equate it to zero.

𝐖𝐫𝐢𝐭𝐢𝐧𝐠 𝐭𝐡𝐞 𝐐𝐮𝐚𝐝𝐫𝐚𝐭𝐢𝐜 𝐅𝐨𝐫𝐦𝐮𝐥𝐚:
\[x=\frac{ -(b) \pm \sqrt{(b)^2-4(a)(c)} }{ 2(a) }\]

Step 1.)
Put the equation in the standard form and find a, b, c:

Step 2.) Substitute a,b, and c into the formula:

Answer 2

\[x=\frac{ -(-3)\pm \sqrt{(-3)^2-4(2)(-5)} }{ 2(2) }\]
Step 3:
Simplify the equation:
\[x=\frac{ 3\pm \sqrt{49}}{ 4 }\]
Step 4: Separate into 2 equations:
\[x=\frac{ 3+ \sqrt{49} }{ 4 }\]
\[x=\frac{ 3- \sqrt{49} }{ 4 }\]
Final Step 5.)
Simplify each equation:
\[x=\frac{ 3+7 }{ 4 }=\frac{ 10 }{ 4 }=2.5\]
\[x=\frac{ 3-7 }{ 4 }=\frac{ -4 }{ 4 }=-1\]