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Solve 2x^2 + 12x = 10. A) -3 +- square root of 14 B) -3 +- 2 of square root of 2 C) -3 +- square root of 19 D) -3 +- square root of 29

User MCardinale
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We will start solving by moving the 10 over by subtraction and set the polynomial equal to 0 so we can factor.
2x^2+12x-10=0. 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.
2(x^2+6x-5)=0 with a=1, b=6, c=-5. Put that into the quadratic formula to get
(-6+/- √(6^2-4(1)(-5)) )/(2) and
(-6+/- √(36+20) )/(2) which simplifies even further to
(-6+/- √(56) )/(2). If we simplify that radical it will simplify to
√(4*14). We can pull the perfect square of 4 out as a 2, leaving us with
2 √(14). So altogether we have
(-6+/-2 √(14) )/(2). The 2 in the denominator will reduce with the other integers (not the radicand!) to give us this as our final answer:
-3+/- √(14), choice A from above.
User VegardKT
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