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3x^2-20x+5=15+9x
We move all terms to the left:
3x^2-20x+5-(15+9x)=0
We add all the numbers together, and all the variables
3x^2-20x-(9x+15)+5=0
We get rid of parentheses
3x^2-20x-9x-15+5=0
We add all the numbers together, and all the variables
3x^2-29x-10=0
a = 3; b = -29; c = -10;
Δ = b2-4ac
Δ = -292-4·3·(-10)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-31}{2*3}=\frac{-2}{6} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+31}{2*3}=\frac{60}{6} =10 $
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