We will start solving by moving the 10 over by subtraction and set the polynomial equal to 0 so we can factor. [tex]2x^2+12x-10=0[/tex]. I am going to factor out the common 2 to make the numbers a bit smaller for when i put them into the quadratic formula. [tex]2(x^2+6x-5)=0[/tex] with a=1, b=6, c=-5. Put that into the quadratic formula to get [tex] \frac{-6+/- \sqrt{6^2-4(1)(-5)} }{2} [/tex] and [tex] \frac{-6+/- \sqrt{36+20} }{2} [/tex] which simplifies even further to [tex] \frac{-6+/- \sqrt{56} }{2} [/tex]. If we simplify that radical it will simplify to [tex] \sqrt{4*14} [/tex]. We can pull the perfect square of 4 out as a 2, leaving us with [tex]2 \sqrt{14} [/tex]. So altogether we have [tex] \frac{-6+/-2 \sqrt{14} }{2} [/tex]. The 2 in the denominator will reduce with the other integers (not the radicand!) to give us this as our final answer: [tex]-3+/- \sqrt{14} [/tex], choice A from above.
