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2x^2=-5x+3
We move all terms to the left:
2x^2-(-5x+3)=0
We get rid of parentheses
2x^2+5x-3=0
a = 2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·2·(-3)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*2}=\frac{-12}{4} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*2}=\frac{2}{4} =1/2 $
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